1 C (CS/IT-Sem-3)
QUANTUM SERIES
For
B.Tech Students of Second Year
of All Engineering Colleges Affiliated to
Dr. A.P.J. Abdul Kalam Technical University,
Uttar Pradesh, Lucknow
(Formerly Uttar Pradesh Technical University)
Discrete Structures & Theory of Logic
By
Dr. Anuj Bhardwaj
M.Sc., Ph.D
Dr. Parul Tiwari
M.Sc., Ph.D
TM
QUANTUM PAGE PVT. LTD.
Ghaziabad
New Delhi
2 C (CS/IT-Sem-3)
PUBLISHED BY :
Apram Singh
Quantum Page Pvt. Ltd.
Plot No. 59/2/7, Site - 4, Industrial Area,
Sahibabad, Ghaziabad-201 010
Phone : 0120 - 4160479
Email : pagequantum@gmail.com
Website: www.quantumpage.co.in
Delhi Office : 1/6590, East Rohtas Nagar, Sahadara, Delhi-110032
© ALL RIGHTS RESERVED
No part of this publication may be reproduced or transmitted,
in any form or by any means, without permission.
Information contained in this work is derived from sources
believed to be reliable. Every effort has been made to ensure
accuracy, however neither the publisher nor the authors
guarantee the accuracy or completeness of any information
published herein, and neither the publisher nor the authors
shall be responsible for any errors, omissions, or damages
arising out of use of this information.
Discrete Structures & Theory of Logic (CS/IT : Sem-3)
1st Edition : 2009-10
12th Edition : 2020-21
nd
2 Edition : 2010-11
3rd Edition : 2011-12
4th Edition : 2012-13
5th Edition : 2013-14
6th Edition : 2014-15
7th Edition : 2015-16
8th Edition : 2016-17
9th Edition : 2017-18
10th Edition : 2018-19
11th Edition : 2019-20 (Thoroughly Revised Edition)
Price: Rs. 80/- only
Printed Version : e-Book.
3 C (CS/IT-Sem-3)
CONTENTS
KCS 303 : Discrete Structures & Theory of Logic
UNIT - 1 : SETS, FUNCTIONS & NATURAL NUMBERS (1–1 C to 1–30 C)
Set Theory: Introduction, Combination of sets, Multisets, Ordered
pairs. Proofs of some general identities on sets. Relations:
Definition, Operations on relations, Properties of relations,
Composite Relations, Equality of relations, Recursive definition of
relation, Order of relations.
Functions: Definition, Classification of functions, Operations on
functions, Recursively defined functions. Growth of Functions.
Natural Numbers: Introduction, Mathematical Induction, Variants
of Induction, Induction with Nonzero Base cases. Proof Methods,
Proof by counter – example, Proof by contradiction.
UNIT - 2 : ALGEBRAIC STRUCTURES
(2–1 C to 2–24 C)
Definition, Groups, Subgroups and order, Cyclic Groups, Cosets,
Lagrange's theorem, Normal Subgroups, Permutation and Symmetric
groups, Group Homomorphisms, Definition and elementary
properties of Rings and Fields.
UNIT - 3 : LATTICES & BOOLEAN ALGEBRA
(3–1 C to 3–22 C)
La tti ces: D efi nition, Properties of la tti ces – Bounded,
Complemented, Modular and Complete lattice.
Boolean Algebra: Introduction, Axioms and Theorems of Boolean
al gebra, Al gebrai c ma nipula tion of Boolea n expressions.
Simplification of Boolean Functions, Karnaugh maps, Logic gates,
Digital circuits and Boolean algebra.
UNIT - 4 : PROPOSITIONAL & PREDICATE LOGIC
(4–1 C to 4–20 C)
Propositional Logic: Proposition, well formed formula, Truth tables,
Tautology, Satisfiability, Contradiction, Algebra of proposition,
Theory of Inference.
Predicate Logic: First order predicate, well formed formula of
predicate, quantifiers, Inference theory of predicate logic.
UNIT - 5 : TREES, GRAPHS & COMBINATORICS
(5–1 C to 5–38 C)
Trees: Definition, Binary tree, Binary tree traversal, Binary search tree.
Graphs: Definition and terminology, Representation of graphs,
Multigraphs, Bipartite graphs, Planar graphs, Isomorphism and
Homeomorphism of graphs, Euler and Hamiltonian paths, Graph
coloring, Recurrence Relation & Generating function: Recursive
definition of functions, Recursive algorithms, Method of solving
recurrences.
Combinatorics: Introduction, Counting Techniques, Pigeonhole
Principle.
SHORT QUESTIONS
SOLVED PAPERS (2014-15 TO 2019-20)
(SQ–1 C to SQ–23 C)
(SP–1 C to SP–24 C)
Discrete Structures & Theory of Logic
1
1–1 C (CS/IT-Sem-3)
Set Theory, Functions
and Natural Numbers
CONTENTS
Part-1
:
Set Theory : Introduction, ........................ 1–2C to 1–5C
Combination of Sets, Multisets
Part-2
:
Ordered Pair, Proof of Some .................... 1–5C to 1–8C
General Identities on Sets
Part-3
:
Relations : Definition, Operation .......... 1–8C to 1–16C
on Relations, Properties of Relation,
Composite Relation, Equality of
Relation, Recursive Definition of
Relation, Order of Relation
Part-4
:
Function : Definition, ............................ 1–16C to 1–22C
Classification of Functions,
Operations on Functions,
Recursively Defined Functions,
Growth of Functions
Part-5
:
Natural Numbers : ................................. 1–22C to 1–26C
Introductions, Mathematical
Induction, Variants of
Induction, Induction with
Non-zero Base Cases
Part-6
:
Proof Methods, Proof by ........................ 1–26C to 1–29C
Counter–Example,
Proof by Contradiction
1–2 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
PART-1
Set Theory : Introduction, Combination of Sets, Multisets.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.1.
What do you understand by set ? Explain different types
of set.
Answer
1.
A set is a collection of well defined objects, called elements or members
of the set.
2. These elements may be anything like numbers, letters of alphabets,
points etc.
3. Sets are denoted by capital letters and their elements by lower case
letters.
4. If an object x is an element of set A, we write it as x A and read it as ‘x
belongs to A’ otherwise x A (x does not belong to A).
Types of set :
1. Finite set : A set with finite or countable number of elements is called
finite set.
2. Infinite set : A set with infinite number of elements is called infinite
set.
3. Null set : A set which contains no element at all is called a null set. It is
denoted by or { }. It is also called empty or void set.
4. Singleton set : A set which has only one element is called singleton set.
5. Subset : Let A and B be two sets, if every elements of A also belongs to
B i.e., if every element of set A is also an element of set B, then A is
called subset of B and it is denoted by A B.
Symbolically, A B if x A x B.
6. Superset : If A is subset of a set B, then B is called superset of A.
7. Proper subset : Any subset A is said to be proper subset of another set
B, if there is at least one element of B which does not belong to A, i.e., if
A B but A B. It is denoted by A B.
8. Universal set : In many applications of sets, all the sets under
consideration are considered as subsets of one particular set. This set is
called universal set and is denoted by U.
9. Equal set : Two set A and B are said to be equal if every element of A
belong to set B and every element of B belong to set A. It is written as
A = B.
Symbolically, A = B if x A and x B.
Discrete Structures & Theory of Logic
1–3 C (CS/IT-Sem-3)
10. Disjoint set : Let A and B be two sets, if there is no common element
between A and B, then they are said to be disjoint.
Que 1.2.
Describe the different types of operation on sets.
Answer
1.
2.
3.
4.
5.
Union : Let A and B be two sets, then the union of sets A and B is a set
that contain those elements that are either in A or B or in both. It is
denoted by A B and is read as ‘A union B’.
Symbolically, A B = {x|x A or x B}
For example :
A = {1, 2, 3, 4 }
B = {3, 4, 5, 6 }
A B = {1, 2, 3, 4, 5, 6}
Intersection : Let A and B be two sets, then intersection of A and B is
a set that contain those elements which are common to both A and B. It
is denoted by A B and is read as ‘A intersection B’.
Symbolically, A B = {x|x A and x B}
For example :
A = {1, 2, 3, 4}
B = {2, 4, 6, 7}
then
A B = {2, 4}
Complement : Let U be the universal set and A be any subset of U,
then complement of A is a set containing elements of U which do not
belong to A. It is denoted by Ac or A or A .
Symbolically,
Ac = {x|x U and x A}
For example :
U = {1, 2, 3, 4, 5, 6}
and
A = {2, 3, 5}
then
Ac = {1, 4, 6}
Difference of sets : Let A and B be two sets. Then difference of A and
B is a set of all those elements which belong to A but do not belong to B
and is denoted by A – B.
Symbolically, A – B = {x|x A and x B}
For example : Let A = {2, 3, 4, 5, 6, 7}
and
B = {4, 5, 7}
then
A – B = {2, 3, 6}
Symmetric difference of set : Let A and B be two sets. The symmetric
difference of A and B is a set containing all the elements that belong to
A or B but not both. It is denoted by A B or A B.
Also
A B = (A B) – (A B)
For example : Let A = {2, 3, 4, 6}
B = {1, 2, 5, 6}
then
A B = {1, 3, 4, 5}
Que 1.3.
What do you mean by multisets ?
Answer
1.
Multisets are sets where an element appear more than once,
1–4 C (CS/IT-Sem-3)
For example :
2.
3.
4.
5.
6.
7.
Set Theory, Functions & Natural Numbers
A = {1, 1, 1, 2, 2, 3}
B = {a, a, a, b, b, b, c, c,}
are multisets.
The multisets A and B can also be written as
A = {3.1, 2.2, 1.3} and B = {3.a, 3.b, 2.c}
The multiplicity of an element in a multiset is defined to be the number
of times an element appears in the multiset. In above examples,
multiplicities of the elements 1, 2, 3 in multiset A are 3, 2, 1 respectively.
Let A and B be two multisets. Then, A B, is the multiset where the
multiplicity of an element is the maximum of its multiplicities in A and B.
The intersection of A and B, A B, is the multiset where the multiplicity
of an element is the minimum of its multiplicities in A and B.
The difference of A and B, A – B, is the multiset where the multiplicity
of an element is equal to the multiplicity of the element in A minus the
multiplicity of the element in B if the difference is positive, and is equal
to zero if the difference is zero and negative.
The sum of A and B, A + B, is the multiset where the multiplicity of an
elements is the sum of multiplicities of the elements in A and B.
Que 1.4.
Let P and Q be two multisets {4.a, 3.b, 1.c} and {3.a, 3.b,
2.d} respectively. Find
i. P Q, ii. P Q, iii. P – Q, iv. Q – P, v. P + Q.
Answer
i.
ii.
iii.
iv.
v.
P Q = {4.a, 3.b, 1.c, 2.d}
P Q = {3.a, 3.b}
P – Q = {1.a, 1.c}
Q – P = {2.d}
P + Q = {7.a, 6.b, 1.c, 2.d}
Que 1.5.
i.
ii.
iii.
iv.
v.
Describe each of following in roster form :
A = {x : x is an even prime}
B = {x : x is a positive integral divisor of 60}
C = {x R : x2 – 1 = 0}
D = {x : x2 – 2x + 1 = 0}
E = {x : x is multiple of 3 or 5}
Answer
i.
ii.
iii.
iv.
v.
A = {2}
B = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}
C = {1, –1}
D = {1}
E = {3, 5, 6, 10, 15 ....}
Que 1.6.
i.
Describe the following in set builder form :
A = {–4, –3, –2, –1, 0, 1, 2, 3}
Discrete Structures & Theory of Logic
1–5 C (CS/IT-Sem-3)
ii. B = {1, 8, 27, 64}
iii. C = {3, 6, 9, 12, 15 ....}
iv. D = {2, 3, 5, 7, 11, 13 .....}
Answer
i.
ii.
iii.
iv.
A = {x|x is an integer and –4 x 3}
B = {x|x = n3, where 1 n 4, n is natural number}
C = {x|x = 3n, where n is natural number}
D = {x|x is the integer and prime number}
PART-2
Ordered Pair, Proof of Some General Identities on Sets.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.7.
List down laws of algebra of sets.
OR
Write down the general identities on sets.
Answer
Let A, B, C be any three sets and U be the universal set, then following are
the laws of algebra of sets :
1. Idempotent laws :
a. A A = A
b. A A = A
2. Commutative laws :
a. A B = B A
b. A B = B A
3. Associative laws :
a. A (B C) = (A B) C
b. A (B C) = (A B) C
4. Distributive laws :
a. A (B C) = (A B) (A C)
b. A (B C) = (A B) (A C)
5. Identity laws :
a. A = A
b. A U = A
c. A U = U
d. A =
6. Involution law :
a. (Ac)c = A
1–6 C (CS/IT-Sem-3)
7.
8.
9.
Set Theory, Functions & Natural Numbers
Complement laws :
a. A Ac = U
b. A Ac =
c. Uc =
d. c = U
De Morgan’s laws :
a. (A B)c = Ac Bc
b. (A B)c = Ac Bc
Absorption laws :
a. A (A B) = A
b. A (A B) = A
Que 1.8.
Prove for any two sets A and B that, (A B) = A B.
AKTU 2014-15, Marks 05
Answer
Let
Now, let
x (A B)
xAB
x A and x B
x A and x B
x A B
(A B) A B
x A B
x A and x B
x A and x B
x (A B)
x (A B)
(AB) (A B)
From eq. (1.8.1) and (1.8.2), (A B) = A B
Que 1.9.
...(1.8.1)
...(1.8.2)
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, and the ordering of
elements of U has the elements in increasing order; that is, ai = i.
What bit strings represent the subset of all odd integers in U, the
subset of all even integers in U, and the subset of integers not
exceeding 5 in U ?
Answer
Since U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let the subset of all odd integers be s1, i.e.,
s1 = {1, 3, 5, 7, 9}
subset of all even integers be s2, i.e.,
s2 = {2, 4, 6, 8, 10}
and the subset of integers not exceeding 5 be s3, i.e.,
s3 = {1, 2, 3, 4, 5}
The bit string by characteristic function is given as follows :
Discrete Structures & Theory of Logic
1–7 C (CS/IT-Sem-3)
Bit string for s1 is 1010101010
Bit string for s2 is 0101010101
Bit string for s3 is 1111100000
Que 1.10. If A and B are two subsets of universal set, then prove
the following :
a. (A – B) = (B – A) iff A = B
b. (A – B) = A iff A B =
Answer
a.
b.
Let
A=B
Consider any element x A – B
x A and x B
x B and x A
[ A = B]
x B–A
A–B B–A
...(1.10.1)
Conversely, if
xB–A
x B and x A
x A and x B
xA–B
B–A A–B
...(1.10.2)
From eq. (1.10.1) and (1.10.2), we have
If
A= BA–B=B–A
Now let
A–B= B–A
Let
x A – B as A – B = B – A
x B–A
Now
x A – B x A and x B
...(1.10.3)
and
x B – A x B and x A
...(1.10.4)
Eq. (1.10.3) and (1.10.4), can hold true when A = B
Let
A–B= A
To show
AB=
Let
A B and let x A B and x
x A and x B
x (A – B) and x B
[ A – B = A]
x A and x B and x B
x
which is a contradiction.
AB=
Now conversely, let A B =
To show A – B = A
Let
xA–B
x A and x B
xA
[as A B = ]
A–B A
...(1.10.5)
Conversely, let x A
x A and x B
[as A B = ]
1–8 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
xA–B
AA–B
From eq. (1.10.5) and (1.10.6),
A=A–B
...(1.10.6)
PART-3
Relations : Definition, Operation on Relations, Properties of
Relation, Composite Relation, Equality of Relation, Recursive
Definition of Relation, Order of Relation.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.11.
Describe the term relation along with its types.
Let A and B be two non-empty sets, then R is relation from A to B if R is
subset of A × B and is set of ordered pair (a, b) where a A and b B. It is
denoted by aRb and read as “a is related to b by R”.
Symbolically, R = {(a, b) : a A, b B, a R b }
If (a, b) R then a R b and read as “a is not related to b by R”.
For example :
Let A = {1, 2, 3, 4}, B = {1, 2} and aRb iff a × b = even number
Then R = {(1, 2), (2, 1), (2, 2), (3, 2), (4, 1), (4, 2)}
Types of relation :
1. Universal relation : A relation R is called universal relation on A if
R = A × A. In case where R is defined from A to B, then R is universal
relation if R = A × B.
For example :
If
A = {1, 2, 3}
then
R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1),
(3, 2), (3, 3)}
is universal relation over A.
2. Identity relation : A relation R is called identity relation on A if
R = {(a, a)|a A}. It is denoted by IA or A or . It is also called diagonal
relation.
For example :
If A = {1, 2, 3}, then
IA = {(1, 1), (2, 2), (3, 3)}
is identity relation on A.
3. Void relation : A relation R is called a void relation on A if R = . It is
also called null relation.
For example :
If A = {1, 2, 3} and R is defined as R = {(a + b) |a + b > 5}, a, b A then
R = .
Discrete Structures & Theory of Logic
4.
5.
1–9 C (CS/IT-Sem-3)
Inverse relation : A relation R defined from B to A is called inverse
relation of R defined from A to B if
R–1 = {(b, a) : b B and a A and (a, b) R}.
For example : Consider relation
R = {(1, 1), (1, 2), (1, 3), (3, 2)}
then
R–1 = {(1, 1), (2, 1), (3, 1), (2, 3)}
Complement of a relation : Let relation R is defined from A to B, then
complement R is set of ordered pairs {(a, b) : (a, b) R}. It is also called
complementary relation.
For example :
Let A = {1, 2, 3}
B = {4, 5}
Then A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Let R be defined as R = {(1, 4), (3, 4), (3, 5)}
Then
Que 1.12.
RC = R = {(1, 5), (2, 4), (2, 5)}
Explain operation on relation with example.
Answer
1.
2.
3.
4.
Relations are sets of ordered pairs so all set operations can be done on
relations.
The resulting sets contain ordered pairs and are, therefore, relations.
If R and S denote two relations, then R S, known as intersection of
R and S, defines a relation such that
x(R S) y = x R y x S y
Similarly, R S, known as union of R and S, such that
x(R S) y = x R y x S y
Also,
x(R – S) y = x R y x $ y where R – S is known as different
of R and S
and
x(R) y = x R y
where R is the complement of R
For example :
A = {x, y, z}, B = {x, y, z}, C = {x, y, z}
D = {Y, z}, R = {(x, X), (x, Y), (y, z)}
S = {(x, Y), (y, z)}
The complement of R consists of all pairs of the Cartesian product
A × B that are not R. Thus A × B = {(x, X), (x, Y), (x, Z), (y, X), (y, Y),
(y, Z), (z, X), (z, Y), (z, Z)}
Hence
R = {(x, Z), (y, X), (y, Y), (z, X), (z, Y), (z, Z)}
R S = {(x, X), (x, Y), (y, Z)}
R S = {(x, Y), (y, Z)}
R – S = {(x, Y)}
Que 1.13.
Give properties of relation.
Answer
Properties of relation are :
1. Reflexive relation : A binary relation R on set A is said to be reflexive
if every element of set A is related to itself.
1–10 C (CS/IT-Sem-3)
2.
3.
4.
5.
6.
7.
i.e.,
a A, (a, a) R or aRa.
For example : Let R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} be a relation
defined on set A = {1, 2, 3}. As (1, 1) R, (2, 2) R and (3, 3) R.
Therefore, R is reflexive relation.
Irreflexive relation : A binary relation R defined on set A is said to
irreflexive if there is no element in A which is related to itself i.e.,
a A such that (a, a) R.
For example : Let R = {(1, 2), (2, 1), (3, 1)} be a relation defined on set
A = {1, 2, 3}. As (1, 1) R, (2, 2) R and (3, 3) R. Therefore, R is
irreflexive relation.
Non-reflexive relation : A relation R defined on set A is said to be nonreflexive if it is neither reflexive nor irreflexive i.e., some elements are
related to itself but there exist at least one element not related to itself.
Symmetric relation : A binary relation on a set A is said to be
symmetric if (a, b) R (b, a) R.
Asymmetric relation : A binary relation on a set A is said to be
asymmetric if (a, b) R (b, a) R.
Antisymmetric relation : A binary relation R defined on a set A is said
to antisymmetric relation if (a, b) R and (b, a) R a = b i.e., aRb and
bRa a = b for a, b R.
Transitive relation : A binary relation R on a set A is transitive
whenever (a, b) R and (b, c) R then (a, c) R
i.e.,
aRb and bRc aRc.
Que 1.14.
a.
b.
Set Theory, Functions & Natural Numbers
Write short notes on :
Equivalence relation
Composition of relation
OR
Write a short note on equality of relation.
Answer
a.
b.
Equivalence relation :
1. A relation R on a set A is said to be equivalence relation if it is
reflexive, symmetric and transitive.
2. The two elements a and b related by an equivalence relation are
called equivalent.
3. So, a relation R is called equivalence relation on set A if it satisfies
following three properties :
i.
(a, a) R a A
(Reflexive)
ii. (a, b) R (b, a) R
(Symmetric)
iii. (a, b) R and (b, c) R (a, c) R
(Transitive)
Composition of relation :
1. Let R be a relation from a set A to B and S be a relation from set B
to C then composition of R and S is a relation consisting or ordered
pair (a, c) where a A and c C provided that there exist b B such
that (a, b) R A × B and (b, c) S B × C. It is denoted by RS.
Discrete Structures & Theory of Logic
2.
1–11 C (CS/IT-Sem-3)
Symbolically, R S = {(a, c)| b B such that (a, b) R and
(b, c) S}
Que 1.15.
Describe recursive definition of relation.
Answer
1.
2.
3.
The characteristic function CR of a relation R Nk is defined as follows :
– CS(X1, ..., Xk) = 1 if <X1, ..., Xk> S
– CS(X1, ..., Xk) = 0 if <X1, ..., Xk> S
A relation R is a recursive set iff its characteristic function CR is a
recursive function.
Examples of recursive relations : <, >, , =
c<(x, y) = sg(y x)
c>(x, y) = sg(x y)
c(x, y) = sg (x y)
4.
5.
6.
7.
c=(x, y) = sg (x y) ×c(x, y) = sg (y x)
Consider relation R(x, y, z) defined as follows :
– R(x, y, z) iff y × z x
We see that R is the result of substituting the recursive function × into
recursive relation .
Thus, R is recursive.
(Technically, R is the result of substituting the functions f1(x, y, z) = y ×
z and f2(x, y, z) = x into , and we need to show that f1(x, y, z) = y × z and
f2(x, y, z) = x are recursive ... but that’s trivial using the identity
functions).
Que 1.16. Define the term partial order relation or partial ordering
relation.
Answer
A binary relation R defined on set A is called Partial Order Relation (POR) if
R satisfies following properties :
i.
(a, a) R a A (Reflexive)
ii. If (a, b) R and (b, a) R, the a = b (Antisymmetric)
iii. If (a, b) R and (b, c) R (a, c) R where a, b, c A (Transitive)
A set A together with a partial order relation R is called partial order set or
poset.
Que 1.17.
a.
b.
c.
Write short notes on :
Closure of relations
Total order
Compatibility relation
1–12 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
Answer
a.
b.
c.
Closure of relations :
i.
Reflexive closure : Let R be a relation defined on set A. The
R IA is called reflexive closure of R, where IA = {(a, a)|a A} is
diagonal or identity relation.
ii. Symmetric closure : Let R be a relation defined on set A. Then
R R–1 is called symmetric closure of R, where R–1 is inverse of R
on A.
Total order : A binary relation R on a set A is said to be total order iff it
is
i.
Partial order
ii. (a, b) R or (b, a) R a, b A
It is also called linear order.
Compatibility relation : A binary relation R defined on set A is said to
be compatible relation if it is reflexive and symmetric. It is denoted by ‘’.
Que 1.18.
Show that R = {(a, b)|a b (mod m)} is an equivalence
relation on Z. Show that if x1 y1 and x2 y2 then (x1 + x2) (y1 + y2).
AKTU 2014-15, Marks 05
Answer
R = {(a, b) | a b (mod m)}
For an equivalence relation it has to be reflexive, symmetric and transitive.
Reflexive : For reflexive a we have (a, a) R i.e.,
a a (mod m)
a – a is divisible by m i.e., 0 is divisible by m
Therefore aRa, a Z, it is reflexive.
Symmetric : Let (a, b) Z and we have
(a, b) R i.e., a b (mod m)
a – b is divisible by m
a – b = km, k is an integer
(b – a) = (– k) m
(b – a) = p m, p is also an integer
b – a is also divisible by m
b a (mod m ) (b, a) R
It is symmetric.
Transitive : Let (a, b) R and (b, c) R then
(a , b) R a – b is divisible by m
a – b = t m, t is an integer
...(1.18.1)
(b, c) R b – c is divisible by m
b – c = s m, s is an integer
...(1.18.2)
From eq. (1.18.1) and (1.18.2)
a – b + b – c = (t + s) m
a – c = lm, l is also an integer
Discrete Structures & Theory of Logic
1–13 C (CS/IT-Sem-3)
a – c is divisible by m
a c (mod m), yes it is transitive.
R is an equivalence relation.
To show : (x1 + x2) (y1 + y2) :
It is given x1 y1 and x2 y2
i.e., x1 – y1 divisible by m
x2 – y2 divisible by m
Adding above equation :
(x1 – y1) + (x2 – y2) is divisible by m
(x1 + x2) – (y1 + y2) is divisible by m
i.e., (x1 + x2) (y1 + y2)
Que 1.19.
Let R be binary relation on the set of all strings of 0’s
and 1’s such that R = {(a, b)|a and b are strings that have the same
number of 0’s}. Is R is an equivalence relation and a partial ordering
relation ?
AKTU 2014-15, Marks 05
Answer
For equivalence relation :
Reflexive : a R a (a, a) R a R
where a is a string of 0’s and 1’s.
Always a is related to a because both a has same number of 0’s.
It is reflexive.
Symmetric : Let (a, b) R
then a and b both have same number of 0’s which indicates that again
both b and a will also have same number of zeros. Hence (b, a) R. It is
symmetric.
Transitive : Let (a, b) R, (b, c) R
(a, b) R a and b have same number of zeros.
(b, c) R b and c have same number of zeros.
Therefore a and c also have same number of zeros, hence (a, c) R.
It is transitive.
R is an equivalence relation.
For partial order, it has to be reflexive, antisymmetric and transitive. Since,
symmetricity and antisymmetricity cannot hold together. Therefore, it is not
partial order relation.
Que 1.20. Let A {1, 2, 3,..........., 13}. Consider the equivalence relation
on A × A defined by (a, b) R (c, d) if a + d = b + c. Find equivalence
classes of (5, 8).
AKTU 2014-15, Marks 05
Answer
A = {1, 2, 3, ...., 13}
[(5, 8)] = [(a, b) : (a, b) R (5, 8), (a, b) A × A]
1–14 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
= [(a, b) : a + 8 = b + 5]
= [(a, b) : a + 3 = b]
[5, 8] = {(1, 4), (2, 5), (3, 6), (4, 7)
(5, 8), (6, 9), (7, 10), (8, 11)
(9, 12), (10, 13)}
Que 1.21.
The following relation on A = {1, 2, 3, 4}. Determine
whether the following :
a. R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}
b. R = A × A
Is an equivalence relation or not ?
AKTU 2015-16, Marks 10
Answer
a.
b.
R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}
Reflexive : (a, a) R a A
(1, 1) R, (2, 2) R
R is not reflexive.
Symmetric : Let (a, b) R then (b, a) R.
(1, 3) R so (3, 1) R
(1, 2) R but (2, 1) R
R is not symmetric.
Transitive : Let (a, b) R and (b, c) R then (a, c) R
(1, 3) R and (3, 1) R so (1, 1) R
(2, 1) R and (1, 3) R but (2, 3) R
R is not transitive.
Since, R is not reflexive, not symmetric, and not transitive so R is not an
equivalence relation.
R=A×A
Since, A × A contains all possible elements of set A. So, R is reflexive,
symmetric and transitive. Hence R is an equivalence relation.
Que 1.22.
Let (A, ) be a partially ordered set. Let be a binary
relation A such that for a and b in A, a is related to b iff b a.
i.
Show that is partially ordered relation.
ii. Show that (A, ) is lattice or not.
Answer
i.
(A, ) is a partially ordered relation if it is reflexive, antisymmetric and
transitive.
Reflexive : Let a A then by definition of relation, aRa a a which
is true.
Hence, the relation R i.e., is reflexive.
Antisymmetric : Let a, b A and
aRb b a
a b
Discrete Structures & Theory of Logic
1–15 C (CS/IT-Sem-3)
b R a
aRb and bRa holds only when a = b. Relation is antisymmetric.
Transitive : Let a, b, c A and aRb and bRc
b a, c b
ca
aRc
Hence, relation is transitive.
Therefore, is a partial order relation.
Since all the elements of the given set A are comparable to each other,
we always have a least upper bound and greatest lower bound for each
pair of elements of A and A is also a partial order relation. Hence, (A, )
is a lattice.
ii.
Que 1.23. Let n be a positive integer and S a set of strings. Suppose
that R n is the relation on S such that sR nt if and only if
s = t, or both s and t have at least n characters and first n characters
of s and t are the same. That is, a string of fewer than n characters is
related only to itself; a string s with at least n characters is related
to a string t if and only if t has at least n characters and t beings with
the n characters at the start of s.
AKTU 2018-19, Marks 07
Answer
We have to show that the relation Rn is reflexive, symmetric, and transitive.
1. Reflexive : The relation Rn is reflexive because s = s, so that sRns
whenever s is a string in S.
2. Symmetric : If sRn t, then either s = t or s and t are both at least n
characters long that begin with the same n characters. This means that
tRn s. We conclude that Rn is symmetric.
3. Transitive : Now suppose that sRn t and tRn u. Then either s = t or s
and t are at least n characters long and s and t begin with the same n
characters, and either t = u or t and u are at least n characters long and
t and u begin with the same n characters. From this, we can deduce that
either s = u or both s and u are n characters long and s and u begin with
the same n characters, i.e., sRn u. Consequently, Rn is transitive.
Que 1.24. Let X = {1, 2, 3,....., 7} and R = {(x, y)|(x – y) is divisible by 3}.
Is R equivalence relation. Draw the digraph of R.
AKTU 2017-18, Marks 07
Answer
Given that
X = {1, 2, 3, 4, 5, 6, 7}
and
R = {(x, y) : (x – y) is divisible by 3 }
Then R is an equivalence relation if
1–16 C (CS/IT-Sem-3)
i.
Set Theory, Functions & Natural Numbers
Reflexive : x X ( x x) is divisible by 3
So, ( x, x) X x X or, R is reflexive.
Symmetric : Let x, y X and (x, y) R
(x – y) is divisible by 3 (x – y) = 3n1, (n1 being an integer)
(y – x) = – 3n2 = 3n2, n2 is also an integer
So, y – x is divisible by 3 or R is symmetric.
iii. Transitive : Let x, y, z X and (x, y) R, (y, z) R
Then x – y = 3n1, y – z = 3n2, n1, n2 being integers
x – z = 3(n1 + n2), n1 + n2 = n3 be any integer
So, (x – z) is also divisible by 3 or (x, z) R
So, R is transitive.
Hence, R is an equivalence relation.
ii.
3
1
4
7
2
6
5
Fig. 1.24.1. Diagraph of R.
PART-4
Function : Definition, Classification of Functions, Operations on
Functions, Recursively Defined Functions, Growth of Functions.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.25.
Define the term function. Also, give classification of it.
Answer
1.
Let X and Y be any two non-empty sets. A function from X to Y is a rule
that assigns to each element x X a unique element y Y.
2. If f is a function from X to Y we write f : X Y.
3. Functions are denoted by f, g, h, i etc.
4. It is also called mapping or transformation or correspondence.
Domain and co-domain of a function : Let f be a function from X to Y.
Then set X is called domain of function f and Y is called co-domain of function f.
1–17 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Range of function : The range of f is set of all images of elements of X.
i.e., Range (f) = {y : y Y and y = f (x) x X }
Also Range (f) Y
Classification of functions :
1. Algebraic functions : Algebraic functions are those functions which
consist of a finite number of terms involving powers and roots of the
independent variable x.
Three particular cases of algebraic functions are :
i.
Polynomial functions : A function of the form a0xn + a1xn–1+...
+ an where n is a positive integer and a0, a1, ...., an are real constants
and a0 0 is called a polynomial of x in degree n, for example
f (x) = 2x3 + 5x2 + 7x – 3 is a polynomial of degree 3.
ii.
Rational functions : A function of the form
f (x )
where f (x) and
g( x )
g(x) are polynomials.
iii. Irrational functions : Functions involving radicals are called
2.
3
irrational functions. f(x) = x + 5 is an example of irrational
function.
Transcendental functions : A function which is not algebraic is called
transcendental function.
i.
Trigonometric functions : Six functions sin x, cos x, tan x, sec x,
cosec x, cot x where the angle x is measured in radian are called
trigonometric functions.
ii. Inverse trigonometric functions : Six functions sin–1 x, cos–1 x,
tan–1 x, cot–1 x, sec–1 x, cosec–1 x, are called inverse trigonometric
functions.
iii. Exponential functions : A function f (x) = ax (a > 0) satisfying the
law a = a and axay = ax+y is called the exponential function.
iv. Logarithm functions : The inverse of the exponential function is
called logarithm function.
Que 1.26.
Give the types / operations on functions.
Answer
1.
One-to-one function (Injective function or injection) : Let
f : X Y then f is called one-to-one function if for distinct elements of X
there are distinct image in Y i.e., f is one-to-one iff
f(x1) = f(x2) implies x1 = x2 x1, x2, X
A
B
f
Fig. 1.26.1. One-to-one.
1–18 C (CS/IT-Sem-3)
2.
Set Theory, Functions & Natural Numbers
Onto function (Surjection or surjective function) : Let f : X Y
then f is called onto function iff for every element y Y there is an
element x X with f(x) = y or f is onto if Range (f) = Y.
X
Y
f
Fig. 1.26.2. Onto.
3.
One-to-one onto function (Bijective function or bijection) : A
function which is both one-to-one and onto is called one-to-one onto
function or bijective function.
X
1
2
3
Y
f
a
b
c
Fig. 1.26.3. One-to-one onto.
4.
Many one function : A function which is not one-to-one is called many
one function i.e., two or more elements in domain have same image in
co-domain i.e.,
If f : X Y then f(x1) = f(x2) x1 x2.
X
1
Y
f
b
2
c
3
d
Fig. 1.26.4. Many one.
5.
6.
Identity function : Let f : X X then f is called identity function if
f (a) = a a X i.e., every element of X is image of itself. It is denoted
by I.
Inverse function (Invertible function) : Let f be a bijective function
from X to Y. The inverse function of f is the function that assigns an
element y Y, a unique element x X such that f(x) = y and inverse of
f denoted by f –1. Therefore if f(x) = y implies f–1 (y) = x.
Que 1.27. Determine whether each of these functions is a bijective
from R to R.
a. f(x) = x2 + 1
c.
f(x) = (x2 + 1)/(x2 + 2)
b.
f(x) = x3
AKTU 2015-16, Marks 15
Discrete Structures & Theory of Logic
1–19 C (CS/IT-Sem-3)
Answer
a.
b.
c.
f(x) = x2 + 1
Let x1, x2 R such that
f(x1) = f(x2 )
x 12 + 1 = x 22 + 1
x12 = x22
x1 = ± x2
Therefore, if x2 = 1 then x1 = ± 1
So, f is not one-to-one.
Hence, f is not bijective.
Let x1, x2 R such that f(x1) = f(x2)
x13 = x23
x1 = x2
f is one-to-one.
Let y R such that
y = x3
x = (y)1/3
For y R a unique x R such that y = f (x)
f is onto.
Hence, f is bijective.
Let x1, x2 R such that f(x1) = f(x2)
x12 1
x2 1
= 22
2
x1 2
x2 2
If
x1 = 1, x2 = – 1 then f(x1) = f(x2)
but
x1 x2
f is not one-to-one.
Hence, f is not bijective.
Que 1.28. If f : A B, g : B C are invertible functions, then show
that g f : A C is invertible and (g o f)–1 = f–1 g–1.
AKTU 2014-15, Marks 05
Answer
If f: A B and g : B C be one-to-one onto functions, then g f is also oneonto and (g f)–1 = f–1 g–1
Proof. Since f is one-to-one, f (x1) = f (x2) x1 = x2 for x1, x2 R
Again since g is one-to-one, g (y1) = g (y2) y1 = y2 for y1, y2 R
Now g f is one-to-one, since (g f) (x1) = (g f) (x2) g [f(x1)] = g [f (x2)]
f (x1) = f(x2 )
[g is one-to-one]
x1 = x2
[f is one-to-one]
Since g is onto, for z C, there exists y B such that g (y) = z. Also f being
onto there exists x A such that f (x) = y. Hence z = g (y)
= g [f (x)] = (g f) (x)
1–20 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
This shows that every element z C has pre-image under gof. So, g f is
onto.
Thus, g f is one-to-one onto function and hence (g f)–1 exists.
By the definition of the composite functions, g f : A C. So,
(g f)–1 : C A.
Also g–1 : C B and f–1 : B A.
Then by the definition of composite functions, f–1 g–1 : C A.
Therefore, the domain of (g f)–1 = the domain of f–1 g–1.
Now
(g f)–1 (z) = x (g f) (x) = z
g (f (x)) = z
g (y) = z where y = f (x)
y = g–1 (z)
f–1 (y) = f–1 (g–1 (z)) = (f–1 g–1) (z)
x = (f–1 g–1) (z)
[f–1 (y) = x]
Thus,
(g f)–1 (z) = (f–1 g–1) (z).
So,
(g f)–1 = f–1 g–1.
Que 1.29.
a.
b.
c.
Explain the following :
Composition of functions
Recursive function
Primitive recursion
Answer
a.
Composition of functions :
If f : X Y and g : Y Z then the composition of f and g is a new function
from X to Z denoted by gof = g(f(x))
X
Y
x
y = f(x)
f
Z
z = g f(x)
g
gf
fg
Fig. 1.29.1. Many one onto.
1.
b.
If f and g are considered to be relations then composition of f and g
was denoted by fog whereas the composition of function f and g is
denoted by gof.
2. Composition of functions is not commutative i.e., fog gof.
3. Composition of functions is associative i.e., fo(goh) = (fog)oh.
where h : X Y, g : Y Z, and f : Z W.
Recursive function (Recursively defined function) :
Sometimes it becomes difficult to define a function explicitly then it may
be easy to define this function in terms of itself. This technique is referred
as recursion.
Discrete Structures & Theory of Logic
1–21 C (CS/IT-Sem-3)
This concept of recursion can be used to define sets, sequences and
algorithms also.
We will define three functions which are called Basis function or Initial
functions that are used in defining other functions by induction.
1. Zero function : (Z : z(x) = 0)
2. Successor function : (S : s(x) = x + 1)
3. Projection function or generalized identity function :
c.
Uin : Uin ( x1 , x2 ,.... xn ) xi )
where superscripts n indicates a number of argument of the function
and subscripts i indicates the value of the function is equal to ith
argument.
Primitive recursion :
A function is called primitive recursion if and only if it can be constructed
from the initial functions and other known primitive recursive functions
by a finite number of operations and recursion only.
For example : Consider the function
f(x, y) = x + y
Now
f(x, y + 1) = x + (y + 1)
= (x + y) + 1
= f(x, y) + 1
Also
f(x, 0) = x + 0 = x
Now
f(x, 0) = x = U11 (x)
.....(1.29.1)
Also
f(x, y + 1) = f(x, y) + 1
= S{f(x, y)} where S is the successor function
= {U33[x, y, f(x, y)]}
...(1.29.2)
If we consider g(x) = U11(x) and h(x, y, z) = S·U33 (x, y, z)
From eq. (1.29.1) and (1.29.2), we get f (x, 0) = g(x)
and f(x, y + 1) = h{x, y, f(x, y)}
Thus, f is obtained from the initial functions U11, U33 and S by applying
compositions once and recursion once. Hence, f is primitive recursive.
Que 1.30.
Write short note on growth of functions.
Answer
1.
2.
3.
We need to approximate the number of steps required to execute any
algorithm because of the difficulty involved in expression or difficulty in
evaluating an expression. We compare one function with another
function to know their rate of growths.
If f and g are two functions we can give the statements like ‘f has same
growth rate as g’ or ‘f has higher growth rate than g’.
Growth rate of function can be defined through notation.
a. -Notation (Same order) : This notation bounds a function to
within constant factors. We say f(n) = g(n) if there exist positive
constants n0, c1 and c2 such that to the right of n0 the value of f(n)
always lies between c1g(n) and c2g(n) inclusive.
1–22 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
c2 g(n)
f(n)
c1 g(n)
n 0 f(n) = (g(n))
n
Fig. 1.30.1.
1.37.1.
b.
O-Notation (Upper bound) : This notation gives an upper bound
for a function to within a constant factor. We write f(n) = O(g(n)) if
there are positive constants n0 and c such that to the right of n0, the
value of f(n) always lies on or below cg(n).
cg(n)
f(n)
n
n0 f(n) = O(g(n))
Fig. 1.30.2.
c.
-Notation (Lower bound) : This notation gives a lower bound
for a function to within a constant factor. We write f(n) = g(n)) if
there are positive constants n0 and c such that to the right of n0, the
value of f(n) always lies on or above cg(n).
f(n)
cg(n)
n
n0 f(n) = (g(n))
Fig. 1.30.3.
PART-5
Natural Numbers : Introduction, Mathematical Induction,
Variants of Induction, Induction with Non-zero Base Cases.
1–23 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.31.
Describe mathematical induction.
Answer
Mathematical induction is a technique of proving a proposition over the
positive integers. It is the most basic method of proof used for proving
statements having a general pattern. A formal statement of principle of
mathematical induction can be stated as follows :
Let S(n) be statement that involve positive integer n = 1, 2, ... then
Step 1 : Verify S(1) is true.
(Inductive base)
Step 2 : Assume that S(k) is true for some arbitrary k.
(Inductive hypothesis)
Step 3 : Verify S(k + 1) is true using basis of inductive hypothesis.
(Inductive step)
Que 1.32.
Prove that n3 + 2n is divisible by 3 using principle of
mathematical induction, where n is natural number.
AKTU 2015-16, Marks 10
Answer
Let S(n) : n3 + 2n is divisible by 3.
Step I : Inductive base : For n = 1
(1)3 + 2.1 = 3 which is divisible by 3
Thus, S(1) is true.
Step II : Inductive hypothesis : Let S(k) is true i.e., k3 + 2k is divisible by
3 holds true.
or k3 + 2k = 3s for s N
Step III : Inductive step : We have to show that S(k + 1) is true
i.e., (k + 1)3 + 2(k + 1) is divisible by 3
Consider (k + 1)3 + 2(k + 1)
= k3 + 1 + 3k2 + 3k + 2k + 2
= (k3 + 2k) + 3(k2 + k + 1)
= 3s + 3l where l = k2 + k + 1 N
= 3(s + l)
Therefore, S(k + 1) is true
Hence by principle of mathematical induction S(n) is true for all
n N.
1–24 C (CS/IT-Sem-3)
Que 1.33.
Set Theory, Functions & Natural Numbers
Prove by induction :
n
1
1
1
=
.
+
+ ... +
1.2 2.3
n ( n + 1) (n + 1)
AKTU 2016-17, Marks 10
Answer
Let the given statement be denoted by S(n).
1. Inductive base : For n = 1
2.
3.
1
1
1
=
11 2
1.2
Hence S (1) is true.
Inductive hypothesis : Assume that S (k) is true i.e.,
1
1
1
k
.....
=
1.2 2.3
k(k 1) k 1
Inductive step : We wish to show that the statement is true for
n = k + 1 i.e.,
1
1
1
k1
.....
=
1.2 2.3
(k 1)(k 2)
k2
Now,
=
1
1
1
1
.....
1.2 2.3
k(k 1) ( k 1)(k 2)
k
1
k2 2k 1
k 1 ( k 1)(k 2) ( k 1)(k 2)
k1
k2
Thus, S(k + 1) is true whenever S(k) is true. By principle of mathematical
induction, S(n) is true for all positive integer n.
=
Que 1.34. Prove by the principle of mathematical induction, that
the sum of finite number of terms of a geometric progression,
a + ar + ar2 + ... arn–1 = a(rn – 1)/(r – 1) if r 1.
AKTU 2014-15, Marks 05
Answer
Basis : True for n = 1 i.e.,
L.H.S = a
R.H.S =
a (r 1)
=a
r1
Therefore, L.H.S. = R.H.S.
Induction : Let it be true for n = k i.e.,
1–25 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
a (r k 1)
r 1
Now we will show that it is true for n = k + 1 using eq. (1.34.1)
i.e., a + ar + ar2 +.... + ark-1 + ark
Using eq. (1.34.1), we get
a + ar + ar2 + .... + ark–1 =
...(1.34.1)
a (r k 1)
+ ark
r 1
ar k a ar k1 ar k
a ( r k 1 1)
=
r 1
r 1
which is R.H.S. for n = k + 1, hence it is true for n = k + 1.
By mathematical induction, it is true for all n.
=
Que 1.35.
Prove that if n is a positive integer, then 133 divides
11n + 1 + 122n – 1.
AKTU 2018-19, Marks 07
Answer
We prove this by induction on n.
Base case : For n = 1, 11n+1 + 122n–1 = 112 + 121 = 133 which is divisible by
133.
Inductive step : Assume that the hypothesis holds for n = k, i.e.,
11k+1 + 122k–1 = 133A for some integer A. Then for n = k + 1,
11n+1 + 122n–1 = 11k+1+1 + 122(k+1)–1
= 11k+2 + 122k+1
= 11 * 11k+1 + 144 * 122k–1
= 11 * 11k+1 + 11 * 122k–1 + 133 * 122k–1
= 11[11k+1 + 122k–1] + 133 * 122k–1
= 11* 133A + 133 * 122k–1
= 133[11A + 122k–1]
Thus if the hypothesis holds for n = k it also holds for n = k + 1. Therefore, the
statement given in the equation is true.
Que 1.36.
4
Prove by mathematical induction
2
n – 4n is divisible by 3 for all n > = 2.
AKTU 2017-18, Marks 07
Answer
Base case : If n = 0, then n4 – 4n2 = 0, which is divisible by 3.
Inductive hypothesis : For some n 0, n4 – 4n2 is divisible by 3.
Inductive step : Assume the inductive hypothesis is true for n. We need to
show that (n + 1)4 – 4(n + 1)2 is divisible by 3. By the inductive hypothesis, we
know that n4 – 4n2 is divisible by 3.
Hence (n + 1)4 – 4(n + 1)2 is divisible by 3 if
(n + 1)4 – 4(n + 1)2 – (n4 –4n2) is divisible by 3.
1–26 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
Now (n + 1)4 – 4(n + 1)2 – (n4 – 4n2)
= n4 + 4n3 + 6n2 + 4n + 1 – 4n2 – 8n – 4 – n4 + 4n2
= 4n3 + 6n2 – 4n – 3,
which is divisible by 3 if 4n 3 – 4n is. Since 4n 3 – 4n = 4n(n + 1)
(n – 1), we see that 4n3 – 4n is always divisible by 3.
Going backwards, we conclude that (n + 1)4 – 4(n + 1)2 is divisible by 3, and
that the inductive hypothesis holds for n + 1.
By the Principle of Mathematical Induction, n4 – 4n2 is divisible by 3, for all
n N.
Que 1.37.
Prove by mathematical induction
3 + 33 + 333 + ............ 3333 = (10n + 1 – 9n – 10)/27
AKTU 2017-18, Marks 07
Answer
3 + 33 + 333 + ............ + 3333 ... = (10n + 1 – 9n – 10)/27
Let given statement be denoted by S(n)
1. Inductive base : For n = 1
(102 9(1) 10)
100 19 81
3=
,3=
=3
27
27
27
3 = 3. Hence S(1) is tree.
2. Inductive hypothesis : Assume that S(k) is true i.e.,
3 + 33 + 333 + ............ + 3333 = (10k + 1 – 9k – 10)/27
3. Inductive steps : We have to show that S(k + 1) is also true i.e.,
3 + 33 + 333 + ............ (10k + 2 – 9(k + 1) – 10)/27
Now, 3 + 33 + ............ + 33 ............ 3
= 3 + 33 + 333 + ............ + 3 ............ 3
= (10k + 1 – 9k – 10)/27 + 3(10k + 1 – 1)/9
= (10k + 1 + 9k – 10 + 9.10k + 1 – 9)/27
= (10k + 1 + 9.10k + 1 – 9k – 8 – 10)/27 = (10k + 2 – 9(k + 1) – 10)/27
Thus S(k + 1) is true whenever S(k) is true. By the principle of
mathematical induction S(n) true for all positive integer n.
PART-6
Proof Methods, Proof by Counter – Example, Proof by Contradiction.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 1.38.
What are the different proof methods ?
Discrete Structures & Theory of Logic
1–27 C (CS/IT-Sem-3)
Answer
Different methods of proof are :
1. Direct proof : In this method, we will assume that hypothesis p is true
from the implication p q. We can proof that q is true by using rules of
interference or some other theorem. This will show p q is true. The
combination of p is true and q is false will never occur.
2. Indirect methods : These are of five types as follows :
a. Proof by contrapositive.
b. Proof by contradiction (Reductio ad absurdum).
c. Proof by exhaustive cases.
d. Proof by cases.
e. Proof by counter example.
Que 1.39.
Explain proof by counter example.
Answer
In this method, we will give an example for which given statement is false.
This example is called counter example.
For example : Prove by counter that every positive integer can be written
as sum of square of three integers.
To look for a counter example we will first write successive positive integers
as the sum of square of three integers.
1 = 12 + 02 + 02
2 = 12 + 02 + 12
3 = 12 + 12 + 12
4 = 22 + 02 + 02
5 = 02 + 12 + 22
6 = 12 + 12 + 22
But there is no way to write 7 as a sum of three squares which gives
counter example. Therefore, given statement is false.
Que 1.40.
Explain proof by contradiction with example.
Answer
In this method, we assume that q is false i.e., ~ q is true. Then by using rules
of inference and other theorems we will show the given statement is true
as well as false i.e., we will reach at a contradiction. Therefore, q must be
true.
For example : Prove that
Let
3 is irrational.
3 is a rational number. Then positive prime integer p and q such that
p
3 =
q
p2
3=
p2 = 3q2
...(1.40.1)
q2
1–28 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
3/p2 (3 divides p2)
3/p
( p is an integer)
p = 3x for some x Z
p2 = 9x2
...(1.40.2)
From eq. (1.40.1) and (1.40.2) we get,
9x2 = 3q2
q2 = 3x2
3/q2 (3 divides q2)
3/q
3 divides p and q which is contradiction to our assumption that p and
q are prime. Therefore,
Que 1.41.
3 is irrational.
Write a short note on the following with example :
i.
Proof by contrapositive.
ii. Proof by exhaustive cases.
iii. Proof by cases.
Answer
i.
ii.
Proof by contrapositive : In this we can prove p q is true by
showing ~ q ~ p is true. It is also called proof by contraposition.
Example : Using method of contraposition if n is integer and 3n + 2 is
even then n is even.
Let p : 3n + 2 is even
q : n is even
Let ~ q is true i.e., n is odd
n = 2k + 1, where k Z
Now,
3n + 2 = 3(2k + 1) + 2
= 2(3k + 2) + 1
= 2m + 1 where m = 3k + 2 Z
(3n + 2) is odd ~ p is true
Hence ~ q ~ p is true.
By method of contraposition p q is true.
Proof by exhaustive cases : Some proofs proceed by exhausting all
the possibilities. We will examine a relatively small number of examples
to prove the theorem. Such proofs are called exhaustive proofs.
Example : Prove that n2 + 1 2n where n is a positive integer and
1 n 4.
We will verify the given inequality n2 + 1 2n for n = 1, 2, 3, 4.
For n = 1; 1 + 1 = 2 which is true
For n = 2; 4 + 1 > 22 which is true
For n = 3; 9 + 1 > 23 which is true
For n = 4; 16 + 1 > 24 which is true
In each of these four cases n2 + 1 2n holds true. Therefore, by method
of exhaustive cases n2 + 1 2n, where n is the positive integer and
1 n 4 is true.
Discrete Structures & Theory of Logic
1–29 C (CS/IT-Sem-3)
iii. Proof by Cases : In this method, we will cover up all the possible
cases that we come across while proving the theorem.
Example : Prove that if x and y are real numbers, then max (x, y)
min (x, y) = x + y.
Case I : If x y, then max (x, y) + min (x, y) = y + x = x + y.
Case II : if x y, then max (x, y) + min (x, y) = x + y.
VERY IMPORTANT QUESTIONS
Following questions are very important. These questions
may be asked in your SESSIONALS as well as
UNIVERSITY EXAMINATION.
Q. 1. Prove for any two sets A and B that, (A B) = A B.
Ans. Refer Q. 1.8.
Q. 2. Show that R = {(a, b)|a b (mod m)} is an equivalence
relation on Z. Show that if x1 y1 and x2 y2 then (x1 + x2)
(y1 + y2).
Ans. Refer Q. 1.18.
Q. 3. Let R be binary relation on the set of all strings of 0’s and
1’s such that R = {(a, b)|a and b are strings that have the
same number of 0’s}. Is R is an equivalence relation and a
partial ordering relation ?
Ans. Refer Q. 1.19.
Q. 4. Let A {1, 2, 3,..........., 13}. Consider the equivalence relation
on A × A defined by (a, b) R (c, d) if a + d = b + c. Find
equivalence classes of (5, 8).
Ans. Refer Q. 1.20.
Q. 5. The following relation on A = {1, 2, 3, 4}. Determine whether
the following :
a. R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}
b. R = A × A
Is an equivalence relation or not ?
Ans. Refer Q. 1.21.
Q. 6. Let n be a positive integer and S a set of strings. Suppose
that Rn is the relation on S such that sRnt if and only if
s = t, or both s and t have at least n characters and first n
characters of s and t are the same. That is, a string of fewer
than n characters is related only to itself; a string s with at
1–30 C (CS/IT-Sem-3)
Set Theory, Functions & Natural Numbers
least n characters is related to a string t if and only if t has
at least n characters and t beings with the n characters at
the start of s.
Ans. Refer Q. 1.23.
Q. 7. Let X = {1, 2, 3,....., 7} and R = {(x, y)|(x – y) is divisible by 3}. Is
R equivalence relation. Draw the digraph of R.
Ans. Refer Q. 1.24.
Q. 8. Determine whether each of these functions is a bijective
from R to R.
a. f(x) = x2 + 1
b. f(x) = x3
c. f(x) = (x2 + 1)/(x2 + 2)
Ans. Refer Q. 1.27.
Q. 9. If f : A B, g : B C are invertible functions, then show
that g f : A C is invertible and (g o f)–1 = f–1 g–1.
Ans. Refer Q. 1.28.
Q. 10. Prove that n3 + 2n is divisible by 3 using principle of
mathematical induction, where n is natural number.
Ans. Refer Q. 1.32.
Q. 11. Prove by induction :
n
1
1
1
=
.
+
+ ... +
1.2 2.3
n ( n + 1) (n + 1)
Ans. Refer Q. 1.33.
Q. 12. Prove by the principle of mathematical induction, that
the sum of finite number of terms of a geometric
progression,
a + ar + ar2 + ... arn–1 = a(rn – 1)/(r – 1) if r 1.
Ans. Refer Q. 1.34.
Q. 13. Prove that if n is a positive integer, then 133 divides
11n + 1 + 122n – 1.
Ans. Refer Q. 1.35.
Q. 14. Prove by mathematical induction, n4 – 4n2 is divisible by 3
for all n > = 2.
Ans. Refer Q. 1.36.
Q. 15. Prove by mathematical induction
3 + 33 + 333 + ............ 3333 = (10n + 1 – 9n – 10)/27
Ans. Refer Q. 1.37.
Discrete Structures & Theory of Logic
2–1 C (CS/IT-Sem-3)
2
Algebraic Structures
CONTENTS
Part-1
:
Algebraic Structures : .............................. 2–2C to 2–10C
Definition, Groups,
Subgroups and Order
Part-2
:
Cyclic Group, Cosets, ............................. 2–10C to 2–16C
Lagrange’s Theorem
Part-3
:
Normal Subgroups, ................................ 2–17C to 2–18C
Permutation and
Symmetric of Groups
Part-4
:
Group Homomorphisms, ....................... 2–19C to 2–22C
Definition and Elementary
Properties of Rings and Fields.
2–2 C (CS/IT-Sem-3)
Algebraic Structures
PART-1
Algebraic Structures : Definition, Groups, Subgroups and Order.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 2.1.
What is algebraic structure ? List properties of algebraic
system.
Answer
Algebraic structure : An algebraic structure is a non-empty set G equipped
with one or more binary operations. Suppose * is a binary operation on G.
Then (G, *) is an algebraic structure.
Properties of algebraic system : Let {S, *, +} be an algebraic structure
where * and + binary operation on S :
1. Closure property : (a * b) S
a, b, S
2. Associative property : (a * b) * c = a * (b * c) a, b, c S
3. Commutative property : (a * b) = (b * a)
a, b S
4.
5.
6.
7.
8.
Identity element : e S such that a * e = a (right identity) a S,
e is called identity element of S with respect to operation *.
Inverse element : For every a S, a–1 S such that
a * a–1 = e = a– 1 * a
here a–1 is called inverse of ‘a’ under operation *.
Cancellation property :
a * b = a * c b = c and b * a = c * a b = c a, b, c S and a 0
Distributive property : a, b, c S
a * (b + c) = (a * b) + (a * c) (right distributive)
(b + c) * a = (b * a) + (c * a) (left distributive)
Idempotent property : An element a S is called idempotent element
with respect to operation * if a * a = a.
Que 2.2.
Write short notes on :
i.
Group
iii. Finite and infinite group
v. Groupoid
ii.
iv.
Abelian group
Order of group
Answer
i.
Group : Let (G, *) be an algebraic structure where * is binary operation
then (G, *) is called a group if following properties are satisfied :
Discrete Structures & Theory of Logic
2–3 C (CS/IT-Sem-3)
1.
2.
3.
a * b G a, b G [closure property]
a * (b * c) = (a * b) * c a, b, c G [associative property]
There exist an element e G such that for any a G
a * e = e * a = e [existence of identity]
4. For every a G, element a–1 G
such that a * a–1 = a–1 * a = e
For example : (Z, +), (R, +), and (Q, +) are all groups.
ii. Abelian group : A group (G, *) is called abelian group or commutative
group if binary operation * is commutative i.e., a * b = b * a a, b G
For example : (Z, +) is an abelian group.
iii. Finite group : A group {G, *} is called a finite group if number of
elements in G are finite. For example G = {0, 1, 2, 3, 4, 5} under 6 is a
finite group.
Infinite group : A group {G, *} is called infinite group if number of
element in G are infinite.
For example, (Z, +) is infinite group.
iv. Order of group : Order of group G is the number of elements in group
G. It is denoted by o (G) or |G|. A group of order 1 has only the identity
element.
v. Groupoid : Let (S, *) be an algebraic structure in which S is a nonempty set and * is a binary operation on S. Thus, S is closed with the
operation *. Such a structure consisting of a non-empty set S and a
binary operation defined in S is called a groupoid.
Que 2.3.
Describe subgroup with example.
Answer
If (G, *) is a group and H G. The (H, *) is said to subgroup of G if
(H, *) is also a group by itself. i.e.,
(1) a * b H a, b H (Closure property)
(2) e H such that a * e = a = e * a a H
Where e is called identity of G.
(3) a – 1 H such that a * a – 1 = e = a– 1 * a a H
For example : The set Q+ of all non-zero +ve rational number is subgroup
of Q – {0}.
Que 2.4.
Show that the set G = {x + y 2 |x, y Q} is a group with
respect to addition.
Answer
i.
Closure :
Let
X = x1 +
2 y1
Y = x2 + 2 y2
where x1, x2, y1, y2 Q and X, Y G
2–4 C (CS/IT-Sem-3)
Then X + Y = (x1 +
ii.
Algebraic Structures
2 y1) + (x2 + 2 y2)
= (x1 + x2) + (y1 + y2) 2
= X1 + 2Y1 G where X1, Y1 Q
Therefore, G is closed under addition [ Sum of two rational numbers is
rational].
Associativity :
Let X, Y and Z G
X = x1 + 2 y1 Y = x2 +
where x1, x2, x3, y1, y2, y3 Q
Consider (X + Y) + Z = (x1 +
2 y1 + x2 +
2 y2 Z = x3 +
2 y2) + (x3 +
2 y 3)
= ((x1 + x2) + (y1 + y2)
2 ) + (x3 +
= (x1 + x2 + x3) + (y1 + y2 + y3) 2
Also
X + (Y + Z) = (x1 +
2 y1) + ((x2 +
2 y3
2 y2) + (x3 +
2 y 1)
...(2.4.1)
2 y3))
= (x1 +
2 y1) + ((x2 + x3) + (y2 + y3) 2 )
= (x1 + x2 + x3) + (y1 + y2 + y3) 2
...(2.4.2)
From eq. (2.4.1) and (2.4.2)
(X + Y) + Z = X + (Y + Z)
Therefore, G is associative under addition.
iii. Identity element :
Let e G be identity elements of G under addition then
(x + y 2 ) + (e1 + e2 2 ) = x + y 2
where e = e1 + e2 2 and e1, e2, x, y Q
e1 + e2 2 = 0 + 0 2
e1 = 0 and e2 = 0
Therefore, 0 G is identity element.
iv. Inverse element :
– x – y 2 G is inverse of x + y 2 G.
Therefore, inverse exist for every element x + y 2 G such that, y Q.
Hence, G is a group under addition.
Que 2.5.
Let H be a subgroup of a finite group G. Prove that order
of H is a divisor of order of G.
AKTU 2018-19, Marks 07
Answer
1.
2.
Let H be any sub-group of order m of a finite group G of order n. Let us
consider the left coset decomposition of G relative to H.
We will show that each coset aH consists of m different elements.
Let
H = {h1, h2, ....., hm}
2–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
3.
4.
Then ah1, ah2, ...., ahm, are the members of aH, all distinct.
For, we have
ahi = ahj hi = hj
by cancellation law in G.
Since G is a finite group, the number of distinct left cosets will also be
finite, say k. Hence the total number of elements of all cosets is km
which is equal to the total number of elements of G.
Hence
n = mk
This show that m, the order of H, is a divisor of n, the order of the
group G.
We also find that the index k is also a divisor of the order of the group.
Que 2.6.
Define identity and zero elements of a set under a binary
operation *. What do you mean by an inverse element ?
Answer
Identity element : An element e in a set S is called an identity element with
respect to the binary operation * if, for any element a in S
a*e= e*a=a
If a * e = a, then e is called the right identity element for the operation * and
if e * a = a, then e is called the left identity element for the operation *.
Zero element : Let R be an abelian group with respect to addition. The
element 0 R will be the additive identity. It is called the zero element of R.
Inverse element : Consider a set S having the identity element e with
respects to the binary operation *. If corresponding to each element
a S there exists an element b S such that
a*b= b*a=e
Then b is said to be the inverse of a and is usually denoted by a – 1. We say a
is invertible.
Que 2.7.
Prove that (Z6, (+6)) is an abelian group of order 6, where
AKTU 2014-15, Marks 05
Z6 = {0, 1, 2, 3, 4, 5}.
Answer
The composition table is :
Since
+6
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
1
2
3
4
5
0
2
3
4
5
0
1
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
2 +6 1 = 3
2–6 C (CS/IT-Sem-3)
Algebraic Structures
4 +6 5 = 3
From the table we get the following observations :
Closure : Since all the entries in the table belong to the given set Z6. Therefore,
Z6 is closed with respect to addition modulo 6.
Associativity : The composition ‘+6’ is associative. If a, b, c are any three
elements of Z6,
a +6 (b +6 c) = a +6 (b + c) [ b +6 c = b + c (mod 6)]
= least non-negative remainder when a + (b + c) is divided by 6.
= least non-negative remainder when (a + b) + c is divided by 6.
= (a + b) +6 c = (a +6 b) +6 c.
Identity : We have 0 Z6. If a is any element of Z6, then from the composition
table we see that
0 +6 a = a = a +6 0
Therefore, 0 is the identity element.
Inverse : From the table we see that the inverse of 0, 1, 2, 3, 4, 5 are 0, 5, 4,
3, 2, 1 respectively. For example 4 +6 2 = 0 = 2 +6 4 implies 4 is the inverse
of 2.
Commutative : The composition is commutative as the elements are
symmetrically arranged about the main diagonal. The number of elements
in the set Z6 is 6.
(Z6, +6) is a finite abelian group of order 6.
Que 2.8.
Let G = {1, – 1, i, – i} with the binary operation
multiplication be an algebraic structure, where i =
1 . Determine
AKTU 2018-19, Marks 07
whether G is an abelian or not.
Answer
The composition table of G is
1.
2.
3.
4.
*
1
–1
i
–i
1
1
–1
i
–i
–1
–1
1
–i
i
i
i
–i
–1
1
–i
–i
i
–1
1
Closure property : Since all the entries of the composition table are
the elements of the given set, the set G is closed under multiplication.
Associativity : The elements of G are complex numbers, and we know
that multiplication of complex numbers is associative.
Identity : Here, 1 is the identity element.
Inverse : From the composition table, we see that the inverse elements
of 1, – 1, i, – i are 1, – 1, – i, i respectively.
2–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
5.
Commutativity : The corresponding rows and columns of the table
are identical. Therefore the binary operation is commutative. Hence,
(G, *) is an abelian group.
Que 2.9.
Write the properties of group. Show that the set (1, 2, 3,
4, 5) is not group under addition and multiplication modulo 6.
AKTU 2017-18, Marks 07
Answer
Properties of group : Refer Q. 2.2(i), Page 2–2C, Unit-2.
Numerical :
Addition modulo 6 (+6) : Composition table of S = {1, 2, 3, 4, 5} under
operation +6 is given as :
+6
1
2
3
4
5
1
2
3
4
2
3
4
5
3
4
5
0
4
5
0
1
5
0
1
2
0
1
2
3
5
0
1
2
3
4
Since, 1 +6 5 = 0 but 0 S i.e., S is not closed under addition modulo 6.
So, S is not a group.
Multiplication modulo 6 (*6) :
Composition table of S = {1, 2, 3, 4, 5} under operation *6 is given as
*6
1
2
3
4
5
1
2
3
4
1
2
3
4
2
4
0
2
3
0
3
0
4
2
0
4
5
4
3
2
5
5
4
3
2
1
Since, 2 *6 3 = 0 but 0 S i.e., S is not closed under multiplication modulo 6.
So, S is not a group.
Que 2.10.
Let G = {a, a2, a3, a4, a5, a6 = e}. Find the order of every
element.
Answer
o(a) : a6 = e o(a) = 6
o(a2) : (a2)3 = a6 = e o (a2) = 3
o(a3) : (a3)2 = a6 = e o(a3) = 2
2–8 C (CS/IT-Sem-3)
Algebraic Structures
o(a4) : (a4)3 = a12 = (a6)2 = e2 = e o(a4) = 3
o(a5) : (a5)6 = a30 = (a6)5 = e5 = e o(a5) = 6
o(a6) : (a6)1 = a6 = e o(a6) = 1
Que 2.11.
Let G be a group and let a, b G be any elements.
Then
i.
(a–1)–1 = a
ii.
(a * b)–1 = b–1 * a–1.
AKTU 2014-15, Marks 05
Answer
i.
ii.
Let e be the identity element for * in G.
Then we have a * a – 1 = e, where a – 1 G.
Also (a – 1) – 1 * a – 1 = e
Therefore, (a – 1) – 1 * a – 1 = a * a – 1.
Thus, by right cancellation law, we have (a – 1) – 1 = a.
Let a and b G and G is a group for *, then a * b G (closure)
Therefore, (a * b) – 1 * (a * b) = e.
....(2.11.1)
Let a – 1 and b – 1 be the inverses of a and b respectively, then a – 1,
b – 1 G.
Therefore, (b – 1 * a – 1) * (a * b) = b – 1 * (a – 1 * a) * b
(associativity)
= b–1 * e * b = b–1 * b = e
...(2.11.2)
From (2.11.1) and (2.11.2) we have,
(a * b) – 1 * (a * b) = (b – 1 * a – 1)* (a * b)
(a * b) – 1 = b – 1 * a –1
(by right cancellation law)
Que 2.12.
Prove that the intersection of two subgroups of a group
is also subgroup.
AKTU 2014-15, Marks 05
Answer
Let H1 and H2 be any two subgroups of G. Since at least the identity element
e is common to both H1 and H2.
H1 H2
In order to prove that H1 H2 is a subgroup, it is sufficient to prove that
a H1 H2, b H1 H2 ab–1 H1 H2
Now a H1 H2 a H1 and a H2
b H1 H2 b H1 and b H2
But H1, H2 are subgroups. Therefore,
a H1, b H1 ab–1 H1
a H2 , bH2 ab–1 H2
Finally, ab–1 H1, ab–1 H2 ab–1 H1 H2
Thus, we have shown that
a H1 H2, b H1 H2 ab–1 H1 H2.
Hence, H1 H2 is a subgroup of G.
2–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Que 2.13.
Let G be the set of all non-zero real number and let
a * b = ab/2. Show that (G *) be an abelian group.
AKTU 2015-16, Marks 10
Answer
i.
Closure property : Let a, b G.
a*b=
ii.
ab
G as ab 0
2
* is closure in G.
Associativity : Let a, b, c G
Consider
bc a(bc) abc
a * (b * c) = a *
2
4
4
ab
(ab) c abc
(a * b) * c = * c =
2
4
4
* is associative in G.
iii. Existence of the identity : Let a G and e such that
ae
=a
2
ae = 2a
e=2
2 is the identity element in G.
iv. Existence of the inverse : Let a G and b G such that a * b = e = 2
ab
=2
2
ab = 4
4
b=
a
4
The inverse of a is , a G.
a
v. Commutative : Let a, b G
a*e=
ab
2
ba ab
and
b*a=
2
2
* is commutative.
Thus, (G, *) is an abelian group.
a*b=
Que 2.14. Prove that inverse of each element in a group is unique.
AKTU 2015-16, Marks 10
2–10 C (CS/IT-Sem-3)
Algebraic Structures
Answer
Let (if possible) b and c be two inverses of element a G.
Then by definition of group :
b*a= a*b=e
and
a*c= c*a=e
where e is the identity element of G
Now
b = e * b = (c * a) * b
= c * (a * b)
= c*c
=c
b=c
Therefore, inverse of an element is unique in (G, *).
PART-2
Cyclic Group, Cosets, Lagrange’s Theorem.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 2.15.
Define cyclic group with suitable example.
Answer
Cyclic group : A group G is called a cyclic group if at least one element a
in G such that every element x G is of the form an, where n is some integer.
The element a G is called the generator of G.
For example :
Show that the multiplicative group G = {1, – 1, i, – i} is cyclic. Also find its
generators.
We have,
i1 = i, i2 = – 1, i3 = – i, i4 = 1.
and
(– i)1 = – i, (– i)2 = – 1, (– i)3 = i, (– i)4 = 1
Thus, every element in G be expressed as in or (– i)n
G is cyclic group and its generators are i and – i.
Que 2.16.
Prove that every group of prime order is cyclic.
AKTU 2018-19, Marks 07
Answer
1.
2.
Let G be a group whose order is a prime p.
Since P > 1, there is an element a G such that a e.
2–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
3.
4.
5.
6.
The group <a> generated by ‘a’ is a subgroup of G.
By Lagrange’s theorem, the order of ‘a’ divides |G|.
But the only divisors of |G|= p are 1 and p. Since a e we have |<a>|> 1,
so |<a>| = p.
Hence, <a> = G and G is cyclic.
Que 2.17.
Show that every group of order 3 is cyclic.
AKTU 2014-15, Marks 05
Answer
1.
2.
3.
4.
5.
6.
7.
Suppose G is a finite group whose order is a prime number p, then to
prove that G is a cyclic group.
An integer p is said to be a prime number if p 0, p 1, and if the only
divisors of p are ± 1, ± p.
Some G is a group of prime order, therefore G must contain at least 2
element. Note that 2 is the least positive prime integer.
Therefore, there must exist an element a G such that a the identity
element e.
Since a is not the identity element, therefore o(a) is definitely 2. Let
o(a) = m. If H is the cyclic subgroup of G generated by a then
o (H = o (a) = m).
By Lagrange’s theorem m must be a divisor of p. But p is prime and
m 2. Hence, m = p.
H = G. Since H is cyclic therefore G is cyclic and a is a generator of G.
Que 2.18.
Show that group (G, +5) is a cyclic group where G = {0, 1,
2, 3, 4}. What are its generators ?
Answer
Composition table of G under operation +5 is shown in Table 2.18.1.
Table 2.18.1.
+5
0
1
2
3
4
0
0
1
2
3
4
1
1
2
3
4
0
2
2
3
4
0
1
3
3
4
0
1
2
4
4
0
1
2
3
0 is the identity element of G under +5.
11 = 1 1 mod 5
12 = 1 + 1 2 mod 5
13 = 1 + 1 + 1 3 mod 5
14 = 1 + 1 + 1 + 1 4 mod 5
15 = 1 = 1 + 1 + 1 + 1 0 mod 5
2–12 C (CS/IT-Sem-3)
Algebraic Structures
where 1n means 1 is added n times
Therefore, 1 is generator of G.
Similarly, 2, 3, 4 are also generators of G.
Therefore, G is a cyclic group with generator 1, 2, 3, 4.
Que 2.19.
Show that G = [(1, 2, 4, 5, 7, 8), X9] is cyclic. How many
generators are there ? What are they ?
AKTU 2015-16, Marks 7.5
Answer
Composition table for X9 is
X9
1
2
4
5
7
8
1
2
4
5
7
8
2
2
4
5
7
8
3
4
8
1
5
7
4
8
7
2
1
5
5
1
2
7
8
4
7
5
1
8
4
2
8
7
5
4
2
1
1 is identity element of group G
21 = 2 2 mod 9
22 = 4 4 mod 9
23 = 8 8 mod 9
24 = 16 7 mod 9
25 = 32 5 mod 9
26 = 64 1 mod 9
Therefore, 2 is generator of G. Hence G is cyclic.
Similarly, 5 is also generator of G.
Hence there are two generators 2 and 5.
Que 2.20.
Define cosets. Write and prove properties of cosets.
Answer
Let H be a subgro up o f gro up G and le t a G the n the se t
Ha = {ha : h H} is called right coset generated by H and a.
Also the set aH = {ah : h H} is called left coset generated by a and H.
Properties of cosets : Let H be a subgroup of G and let a and b belong to
G. Then
1. a a H
Proof : a = a e a H
Since e is identity element of G.
2. aH = H iff a H.
Proof : Let aH = H.
Then a = ae aH = H (e is identity in G and so is in H)
aH
3.
aH = bH or aH bH =
Discrete Structures & Theory of Logic
2–13 C (CS/IT-Sem-3)
Proof : Let aH = bH or aH bH =
and to prove that aH = bH.
Let aH bH
Then there exists h1, h2 H such that
x = ah1 and x = bh2
a = xh1–1 = bh2 h1–1
Since H is a subgroup, we have h2 h1–1 H
let h2 h1–1 = h H
Now, aH = bh2h1–1 H = (bh) H = b (hH) = bh ( hH = H by property 2)
aH = bH if a H bH
4.
5.
Thus, either a H bH = or aH = bH.
aH = bH iff a–1 b H.
Proof : Let
aH = bH.
a–1 aH = a–1 bH
eH = a–1 bH
(e is identity in G)
H = (a–1 b) H
Therefore by property (2); a–1 b H.
Conversely, now if a–1 b H.
Then consider bH = e (bH) = (a a–1) (bH)
= a (1–1 b) H
= aH
Thus
aH = bH iff a–1 b H.
aH is a subgroup of G iff a H.
Proof : Let aH is a subgroup of G then it contains the identity e of G.
Thus, aH eH
then by property (3) ; aH = eH = H
aH = a H
Conversely, if a H then by property (2); aH = H.
Que 2.21.
State and prove Lagrange’s theorem for group. Is the
converse true ?
AKTU 2016-17, Marks 10
Answer
Lagrange’s theorem :
Statement : The order of each subgroup of a finite group is a divisor of the
order of the group.
Proof : Let G be a group of finite order n. Let H be a subgroup of G and let
O(H) = m. Suppose h1, h2.... ..., hm are the m members of H.
Let a G, then Ha is the right coset of H in G and we have
Ha = {h1 a, h2 a,....hm a}
Ha has m distinct members, since = hi a = hj a hi = hj
Therefore, each right coset of H in G has m distinct members. Any two
distinct right cosets of H in G are disjoint i.e., they have no element in
2–14 C (CS/IT-Sem-3)
Algebraic Structures
common. Since G is a finite group, the number of distinct right cosets of H in
G will be finite, say, equal to k. The union of these k distinct right cosets of H
in G is equal to G.
Thus, if Ha1, Ha2,...., Hak are the k distinct right cosets of H in G. Then
G = Ha1 Ha2 Ha3 .... Hak
the numbe r of elements in G = the numbe r of e lements in
Ha1 + ...... + the number of elements in Ha2 +......+ the number of elements in
Hak
O(G) = km
n = km
n
k=
m
m is a divisor of n.
O(H) is a divisor of O(G).
Proof of converse : If G be a finite group of order n and n G, then
an = e
Let o (a) = m which implies am = e.
Now, the subset H of G consisting of all the integral power of a is a subgroup
of G and the order of H is m.
Then, by the Lagrange’s theorem, m is divisor of n.
Let n = mk, then
an = amk = (am)k = ek = e
Yes, the converse is true.
Que 2.22.
State and explain Lagrange’s theorem.
Answer
Lagrange’s theorem :
If G is a finite group and H is a subgroup of G then o(H) divides o(G).
Moreover, the number of distinct left (right) cosets of H in G is o(G)/o(H).
Proof : Let H be subgroup of order m of a finite group G of order n.
Let H {h1, h2, ..., hm}
Let a G. Then aH is a left coset of H in G and aH = {ah1, ah2, ..., ahm} has m
distinct elements as ahi = ahj hi = hj by cancellation law in G.
Thus, every left coset of H in G has m distinct elements.
Since G is a finite group, the number of distinct left cosets will also be finite.
Let it be k. Then the union of these k-left cosets of H in G is equal to G.
i.e., if a1H, a2H, ..., akH are right cosets of H in G then
G = a1H a2H ... akH.
o(G) = o(a1H) + o(a2H) + ... + o(akH)
(Since two distinct left cosets are mutually disjoint.)
n = m + m + ... + m (k times)
n
n = mk k =
m
o(G )
k=
.
o( H )
Discrete Structures & Theory of Logic
2–15 C (CS/IT-Sem-3)
Thus order of each subgroup of a finite group G is a divisor of the order of the
group.
Cor 1 : If H has m different cosets in G then by Lagrange’s theorem :
o(G) = m o(H)
m=
o(G )
o( H )
o(G )
o( H )
Cor 2 : If |G| = n and a G then an = e
Let |a| = m am = e
Now, the subset H of G consisting of all integral powers of a is a subgroup of
G and the order of H is m.
Then by Lagrange’s theorem, m is divisor of n.
Let
n = mk, then
an = amk = (am)k = ek = e
[G : H] =
Que 2.23.
a.
b.
c.
Prove that every cyclic group is an abelian group.
Obtain all distinct left cosets of {(0), (3)} in the group
(Z6, +6) and find their union.
Find the left cosets of {[0], [3]} in the group (Z6, +6).
AKTU 2016-17, Marks 10
Answer
a.
Let G be a cyclic group and let a be a generator of G so that
G = <a> = {an : n Z}
If g1 and g2 are any two elements of G, there exist integers r and s such
that g1 = ar and g2 = as. Then
g1 g2 = ar as = ar+s = as+r = as . ar = g2g1
So, G is abelian.
b. [0] + H = [3] + H, [1] + [4] + H and [2] + H = [5] + H are the three
distinct left cosets of H in (Z6 , + 6 ).
We would have the following left cosets :
g1H = {g1 h, h H}
g2H = {g2 h, h H}
gnH = {gn h, h H}
The union of all these sets will include all the g s, since for each set
gk = {gk h, h H}
we have
ge gk = {gk h, h H}
where e is the identity.
Then if we make the union of all these sets we will have at least all the
elements of g. The other elements are merely gh for some h. But since ghG
they would be repeated elements in the union. So, the union of all left cosets
of H in G is G, i.e.,
Z6 = {[0], [1], [2], [3], [4], [5]}
2–16 C (CS/IT-Sem-3)
c.
Algebraic Structures
Let
Z6 = {[0], [1], [2], [3], [4], [5]} be a group.
H = {[0], [3]} be a subgroup of (Z6, +6).
The left cosets of H are,
[0] + H = {[0], [3]}
[1] + H = {[1], [4]}
[2] + H = {[2], [5]}
[3] + H = {[3], [0]}
[4] + H = {[4], [1]}
[5] + H = {[5], [2]}
Que 2.24.
Write and prove the Lagrange’s theorem. If a group
G = {...., – 3, 2, – 1, 0, 1, 2, 3,.....} having the addition as binary operation.
If H is a subgroup of group G where x2 H such that x G. What is
H and its left coset w.r.t 1 ?
AKTU 2014-15, Marks 05
Answer
Lagrange’s theorem : Refer Q. 2.22, Page 2–14C, Unit-2.
Numerical :
H = {x2 : x G} = {0, 1, 4, 9, 16, 25 ....}
Left coset of H will be 1 + H = {1, 2, 5, 10, 17, 26, ....}
Que 2.25. What do you mean by cosets of subgroup ? Consider the
group Z of integers under addition and the s ubgroup
H = {..., –10, –5, 0, 5, 10, ....} considering the multiple of 5.
i.
Find the cosets of H in Z.
ii. What is index of H in Z ?
Answer
Coset : Refer Q. 2.20, Page 2–12C, Unit-2.
Numerical :
i.
We have Z = {– 7, – 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6, ....}
and
H = {...., – 10, – 5, 0, 5, 10, ....}
Let 0 Z and the right cosets are given as
H = H + 0 = {...., – 10, – 5, 0, 5, 10, ....}
H + 1 = {...., – 9, – 4, 1, 6, 11, ....}
H + 2 = {...., – 8, – 3, 2, 7, 12, ....}
H + 3 = {...., – 7, – 2, 3, 8, 13, ....}
H + 4 = {....., – 6, – 1, 4, 9, 14, ....}
H + 5 = {...., – 10, – 5, 0, 5, 10, ....] = H
Now, its repetition starts. Now, we see that the right cosets,
H, H + 1, H + 2, H + 3, H + 4 are all distinct and more over they are
disjoint. Similarly the left cosets will be same as right cosets.
ii. Index of H in G is the number of distinct right/left cosets.
Therefore, index is 5.
Discrete Structures & Theory of Logic
2–17 C (CS/IT-Sem-3)
PART-3
Normal Subgroups, Permutation and Symmetric of Groups.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 2.26.
a.
b.
Write short notes on :
Normal subgroup
Permutation group
Answer
a.
b.
Normal subgroup : A subgroup H of G is said to be normal subgroup
of G if Ha = aH a G i.e., the right coset and left coset of H is G
generated by a are the same.
i.
Clearly, every subgroup H of an abelian group G is a normal
subgroup of G. For a G and h H, ah = ha.
ii. Since a cyclic group is abelian, every subgroup of a cyclic group is
normal.
Permutation group : A set G of all permutation on a non-empty
set A under the binary operation * is called permutation group.
If A = {1, 2, 3, ...., n}, the given permutation group formed by A is also
called symmetric group of degree n denoted by Sn. Number of elements
of Sn will be n!.
Cyclic permutation : Let A = {x1, x2, ..., xn}. Then let t1, t2,..., tk be
elements of set A and permutation P : A defined by
P(t1) = t2
P(t2) = t3
...........
...........
P(tk–1) = tk
P(tk) = t1
is called a cyclic permutation of length k.
a b c
For example : Consider A = {a, b, c, d, e}. Then let P =
c b d
Then P has a cycle of length 3 given by (a, c, d).
Que 2.27.
d e
.
a e
Define the subgroup of a group. Let (G, ) be a group.
Let H = {a | a G and a b = b a for all b G}. Show that H is a
normal subgroup.
2–18 C (CS/IT-Sem-3)
Algebraic Structures
Answer
Subgroup : If (G, *) is a group and H G. Then (H, *) is said to subgroup of
G if (H, *) is also a group by itself.
i.e.,
1. a * b H a, b H (Closure property)
2. e H such that a * e = a = e * a a H
where e is called identity of G.
3. a–1 H such that a * a–1 = e = a–1 * a a H
Numerical : Let (G, x) be a group. A non-empty subset H of a group G is said
to be a subgroup of G if (H, *) itself is a group.
Given that,
H = {a|a G and a b = b a; b G}
Let
a, b H a x = x a and b x = x b, x G
(b x) – 1 = (x b) – 1
x – 1 b – 1 = b – 1 x1
b – 1H.
Now,
(a b – 1) x = a (b – 1 x)
[ is associative]
= a (x b – 1)
[ use b – 1 H]
= (a x) b – 1
[ a H]
= (x a) b – 1
= x (a b – 1)
–
1
ab H
Therefore, H is a subgroup of group G.
Let h H and g G and any x in G.
Consider
(g h g – 1) x = (g g – 1 h) x
[ h H]
= (e h) x = h x
= xh
[ h H]
= x (h g g – 1)
= x (g h g – 1)
[ h H]
–
1
g h g H for any g G
H is a normal subgroup of G.
Que 2.28.
If N and M are normal subgroup of G then N M is a
normal subgroup of G.
Answer
As N and M are subgroups of G then N M is a subgroup of G. Let g G
and a N M
a N and a M.
Since N is normal subgroup of G, gag–1 N
Since M is normal subgroup of G, gag–1 M
gag–1 N M is a normal subgroup of G.
Hence N M is a normal subgroup of G.
Discrete Structures & Theory of Logic
2–19 C (CS/IT-Sem-3)
PART-4
Group Homomorphism, Definition and Elementary Properties of
Ring and Fields.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 2.29.
Discuss homomorphism and isomorphic group.
Answer
Homomorphism : Let (G1, •) and (G2, *) be two groups then a mapping
f : G1 G2 is called a homomorphism if f(a • b) = f(a) * f(b) for all a, b G1.
Thus f is homomorphism from G1 to G2 then f preserves the composition in
G1 and G2 i.e., image of composition is equal to composition of images.
The group G2 is said to be homomorphic image of group G1 if there exist a
homomorphism of G1 onto G2.
Isomorphism : Let (G1, •) and (G2, *) be two groups then a mapping
f : G1 G2 is an isomorphism if
i.
f is homomorphism.
ii.
f is one to one i.e., f(x) = f(y) x = y x, y G1.
f is onto.
iii.
Que 2.30. Give the definitions of rings, integral domains and fields.
Answer
Ring : A ring is an algebraic system (R, +, •) where R is a non-empty set and
+ and • are two binary operations (which can be different from addition and
multiplication) and if the following conditions are satisfied :
1.
2.
(R, +) is an abelian group.
(R, •) is semigroup i.e., (a • b) • c = a • (b • c) a, b, c R.
3.
The operation • is distributive over +.
i.e., for any a, b, c R
a • (b + c) = (a • b) + (a • c) or (b + c) • a = (b • a) + (c • a)
Integral domain : A ring is called an integral domain if :
i.
It is commutative
ii.
iii.
It has unit element
It is without zero divisors
2–20 C (CS/IT-Sem-3)
Algebraic Structures
Field : A ring R with at least two elements is called a field if it has following
properties :
i.
R is commutative
ii. R has unity
iii. R is such that each non-zero element possesses multiplicative inverse.
For example, the rings of real numbers and complex numbers are also fields.
Que 2.31.
Consider a ring (R, +, ) defined by a • a = a, determine
whether the ring is commutative or not.
AKTU 2014-15, Marks 05
Answer
Let a, b R (a + b)2 = (a + b)
(a + b) (a + b) = (a + b)
(a + b)a + (a + b)b = (a + b)
(a2 + ba) + (ab + b2) = (a + b)
(a + ba) + (ab + b) = (a + b)
( a2 = a and b2 = b)
(a + b) + (ba + ab) = (a + b) + 0
ba + ab = 0
a + b = 0 a + b = a + a [being every element of its own additive inverse]
b=a
ab = ba
R is commutative ring.
Que 2.32.
Write out the operation table for [Z2, +2, * 2]. Is Z2 a
ring ? Is an integral domain ? Is a field ? Explain.
Answer
The operation tables are as follows :
we have Z2 = {0, 1}
+2
0
1
*2
0
1
0
0
1
0
0
0
1
1
0
1
0
1
Since (Z2, +2, *2) satisfies the following properties :
i.
Closure axiom : All the entries in both the tables belong to Z2. Hence,
closure is satisfied.
ii.
Commutative : In both the tables all the entries about the main diagonal
are same therefore commutativity is satisfied.
iii. Associative law : The associative law for addition and multiplication
are also satisfied.
2–21 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
iv. Here 0 is the additive identity and 1 is the multiplicative identity. Identity
property is satisfied.
v.
Inverse exists in both the tables. The additive inverse of 0, 1 are 1, 0
respectively and the multiplicative inverse of non-zero element of Z2
is 1.
vi. Multiplication is distributive over addition.
Hence (Z2, +2, *2) is a ring as well as field. Since, we know that every field is
an integral domain therefore it is also an integral domain.
Que 2.33.
If the permutation of the elements of {1, 2, 3, 4, 5} are
given by a = (1 2 3) (4 5), b = (1) (2) (3) (4 5), c = (1 5 2 4) (3). Find the value
of x, if ax = b. And also prove that the set Z4 = (0, 1, 2, 3) is a commutative
ring with respect to the binary modulo operation +4 and *4.
AKTU 2015-16, Marks 10
Answer
ax = b
(123) (45) x = (1) (2) (3) (4, 5)
1 2 3 4 5 1 2 3 4 5
1 2 3 4 5
2 3 1 4 5 1 2 3 5 4 x 1 2 3 5 4
1 2 3 4 5
1 2 3 4 5
x=
2 3 1 5 4
1 2 3 5 4
1 2 3 4 5
x=
2 3 1 5 4
1
1 2 3 4 5
1 2 3 5 4
2 3 1 5 4 1 2 3 4 5
=
1 2 3 4 5 1 2 3 5 4
1 2 3 4 5 1 2 3 4 5
=
3 1 2 5 4 1 2 3 5 4
x=
4
0
1
0 1 2 3
0 1 2 3
1 2 3 0
1 2 3 4 5
3 1 2 4 5
4
0
1
2
3
0
0
0
0
0
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
2 2 3 0 1
3 3 0 1 2
We find from these tables :
i.
All the entries in both the tables belong to Z4. Hence, Z4 is closed with
respect to both operations.
2–22 C (CS/IT-Sem-3)
Algebraic Structures
Commutative law : The entries of 1st, 2nd, 3rd, 4th rows are identical
with the corresponding elements of the 1 st , 2nd, 3rd, 4th columns
respectively in both the tables. Hence, Z4 is commutative with respect to
both operations.
iii. Associative law : The associative law for addition and multiplication
a +4 (b +4 c) = (a +4 b) +4 c for all a, b, c Z4
a ×4 (b ×4 c) = (a ×4 b) ×4 c, for all a, b, c Z4
can easily be verified.
iv. Existence of identity : 0 is the additive identity and 1 is multiplicative
identity for Z4.
v. Existence of inverse : The additive inverses of 0, 1, 2, 3 are 0, 3, 2, 1
respectively.
Multiplicative inverse of non-zero element 1, 2, 3 are 1, 2, 3 respectively.
vi. Distributive law : Multiplication is distributive over addition i.e.,
a ×4(b +4c) = a ×4b + a ×4c
(b +4c) ×4a = b ×4a + c ×4a
For,
a ×4(b +4c) = a ×4(b + c) for b +4 c = b + c (mod 4)
= least positive remainder when a × (b + c) is
divided by 4
= least positive remainder when ab + ac is divided
by 4
= ab +4ac
= a ×4b +4a ×4c
For
a ×4b = a × b (mod 4)
Since (Z4, +4) is an abelian group, (Z4, ×4) is a semigroup and the operation is
distributive over addition. The (Z4, +4, ×4) is a ring. Now (Z4, ×4) is commutative
with respect to ×4. Therefore, it is a commutative ring.
ii.
Que 2.34.
What is meant by ring ? Give examples of both
commutative and non-commutative rings.
AKTU 2018-19, Marks 07
Answer
Ring : Refer Q. 2.30, Page 2–19C, Unit-2.
Example of commutative ring : Refer Q. 2.31, Page 2–20C, Unit-2.
Example of non-commutative ring : Consider the set R of 2 × 2 matrix
with real element. For A, B, C R
A * (B + C) = (A * B) + (A * C)
also,
(A + B) * C = (A * C) + (B * C)
* is distributive over +.
(R, +, *) is a ring.
We know that AB BA, Hence (R, +, *) is non-commutative ring.
Discrete Structures & Theory of Logic
2–23 C (CS/IT-Sem-3)
VERY IMPORTANT QUESTIONS
Following questions are very important. These questions
may be asked in your SESSIONALS as well as
UNIVERSITY EXAMINATION.
Q. 1. Let H be a subgroup of a finite group G. Prove that order of
H is a divisor of order of G.
Ans. Refer Q. 2.5.
Q. 2. Prove that (Z6, (+6)) is an abelian group of order 6, where
Z6 = {0, 1, 2, 3, 4, 5}.
Ans. Refer Q. 2.7.
Q. 3. Let G = {1, – 1, i, – i} with the binary operation multiplication
be an algebraic structure, where i =
whether G is an abelian or not.
Ans. Refer Q. 2.8.
1 . Determine
Q. 4. Write the properties of group. Show that the set (1, 2, 3, 4, 5)
is not group under addition and multiplication modulo 6.
Ans. Refer Q. 2.9.
Q. 5. Let G be a group and let a, b G be any elements.
Then
i. (a–1)–1 = a
ii. (a * b)–1 = b–1 * a–1.
Ans. Refer Q. 2.11.
Q. 6. Prove that the intersection of two subgroups of a group is
also subgroup.
Ans. Refer Q. 2.12.
Q. 7. Let G be the set of all non-zero real number and let
a * b = ab/2. Show that (G *) be an abelian group.
Ans. Refer Q. 2.13.
Q. 8. Prove that inverse of each element in a group is unique.
Ans. Refer Q. 2.14.
Q. 9. Prove that every group of prime order is cyclic.
Ans. Refer Q. 2.16.
2–24 C (CS/IT-Sem-3)
Algebraic Structures
Q. 10. Show that every group of order 3 is cyclic.
Ans. Refer Q. 2.17.
Q. 11. Show that G = [(1, 2, 4, 5, 7, 8), X9] is cyclic. How many
generators are there ? What are they ?
Ans. Refer Q. 2.19.
Q. 12. State and prove Lagrange’s theorem for group. Is the
converse true ?
Ans. Refer Q. 2.21.
Q. 13.
a. Prove that every cyclic group is an abelian group.
b. Obtain all distinct left cosets of {(0), (3)} in the group
(Z6, +6) and find their union.
c. Find the left cosets of {[0], [3]} in the group (Z6, +6).
Ans. Refer Q. 2.23.
Q. 14. Write and prove the Lagrange’s theorem. If a group
G = {...., – 3, 2, – 1, 0, 1, 2, 3,.....} having the addition as binary
operation. If H is a subgroup of group G where x2 H such
that x G. What is H and its left coset w.r.t 1 ?
Ans. Refer Q. 2.24.
Q. 15. Consider a ring (R, +, ) defined by a • a = a, determine
whether the ring is commutative or not.
Ans. Refer Q. 2.31.
Q. 16. If the permutation of the elements of {1, 2, 3, 4, 5} are given
by a = (1 2 3) (4 5), b = (1) (2) (3) (4 5), c = (1 5 2 4) (3). Find the
value of x, if ax = b. And also prove that the set Z4 = (0, 1, 2, 3)
is a commutative ring with respect to the binary modulo
operation +4 and *4.
Ans. Refer Q. 2.33.
Q. 17. What is meant by ring ? Give examples of both commutative
and non-commutative rings.
Ans. Refer Q. 2.34.
3–1 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
3
Lattices and
Boolean Algebra
CONTENTS
Part-1
:
Lattices : Definition, ................................. 3–2C to 3–10C
Properties of Lattices–Bounded,
Complemented, Modular
and Complete Lattice
Part-2
:
Boolean Algebra : .................................... 3–10C to 3–13C
Introduction, Axioms and
Theorems of Boolean Algebra
Part-3
:
Algebraic Manipulation .......................... 3–13C to 3–17C
of Boolean Expressions,
Simplification of Boolean
Functions, Karnaugh Maps
Part-4
:
Logic Gates, Digital ................................ 3–17C to 3–20C
Circuits and
Boolean Algebra
3–2 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
PART-1
Lattices : Definition, Properties of Lattices–Bounded,
Complemented, Modular and Complete Lattice.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 3.1.
Define lattice. Give its properties.
Answer
A lattice is a poset (L, ) in which every subset {a, b} consisting of 2 elements
has least upper bound (lub) and greatest lower bound (glb). Least upper
bound of {a, b} is denoted by a b and is known as join of a and b. Greatest
lower bound of {a, b} is denoted by a b and is known as meet of a and b.
Lattice is generally denoted by (L, , ).
Properties :
Let L be a lattice and a, b L then
1. Idempotent property :
i.
aa=a
ii. a a = a
2. Commutative property :
i.
ab=ba
ii. a b = b a
3. Associative property :
i.
a (b c) = (a b) c
ii. a (b c) = (a b) c
4. Absorption property :
i.
a (a b) = a
ii. a (a b) = a
5. i.
a b = b iff a b
ii. a b = a iff a b
iii. a b = a iff a b = b
6. Distributive Inequality :
i.
a (b c) (a b) (a c) ii. a (b c) (a b) (a c)
7.
8.
Isotonicity :
Let a, b, c, L and a b, then
i.
acbc
ii. a c b c
Stability :
Let a, b, c, L, then
i.
a b, a c a (b c), a (b c)
ii. a b, a c a (b c), a (b c)
Que 3.2.
Explain types of lattice.
3–3 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Answer
Types of lattice :
1. Bounded lattice : A lattice L is said to be bounded if it has a greatest
element 1 and a least element 0. In such lattice we have
a 1 = 1, a 1 = a
a 0 = a, a 0 = 0
a L and 0 a I
2. Complemented lattice : Let L be a bounded lattice with greatest
element 1 and least element 0. Let a L then an element a L is
complement of a if,
a a = 1 and a a = 0
A lattice L is called complemented if is bounded and if every element in
L has a complement.
3. Distributive lattice : A lattice L is said to be distributive if for any
element a, b and c of L following properties are satisfied :
i.
a (b c) = (a b) (a c)
ii. a (b c) = (a b) (a c)
otherwise L is non-distributive lattice.
4. Complete lattice : A lattice L is called complete if each of its nonempty subsets has a least upper bound and greatest lower bound.
For example :
i.
(Z, ) is not a complete lattice.
ii. Let S be the class of all subsets of some universal set A and a
relation is defined as X Y X is a subset of Y such that
X Y = X Y and X Y. Every subset of S has glb and lub. So, S
is complete.
5. Modular lattice : A lattice (L, ) is called modular lattice if,
a (b c) = (a b) c whenever a c for all a, b, c L.
Que 3.3.
If the lattice is represented by the Hasse diagram given
below :
i.
Find all the complements of ‘e’.
ii. Prove that the given lattice is bounded complemented lattice.
b
c
a
f
d
e
Fig. 3.3.1.
AKTU 2014-15, Marks 10
3–4 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
Answer
i.
ii.
Complements of e are c and d which are as follows :
c e = b , c e = f
de=b , de=f
A lattice is bounded if it has greatest and least elements. Here b is
greatest and f is least element.
Que 3.4.
Define a lattice. For any a, b, c, d in a lattice (A, ) if a b
and c d then show that a c b d and a c b d.
AKTU 2018-19, Marks 07
Answer
Lattice : Refer Q. 3.1, Page 3–2C, Unit-3.
Numerical :
As a b and c d, a b b d and c d b d.
By transtivity of , a b d and c b d.
So b d is an upper bound of a and c.
So a c bd.
As a c a and a c c, a c a b and a c c d.
Hence a c is a lower bound of b and d. So a c b d.
So a c b d.
Que 3.5.
Let L be a bounded distributed lattice, prove if a
complement exists, it is unique. Is D12 a complemented lattice ?
Draw the Hasse diagram of [P (a, b, c), <], (Note : ‘<’ stands for subset).
Find greatest element, least element, minimal element and maximal
AKTU 2015-16, Marks 15
element.
OR
Draw the Hasse diagram of [P (a, b, c), ] (Note : ‘’ stands for subset).
Find greatest element, least element, minimal element and maximal
AKTU 2015-16, Marks 10
element.
Answer
Let a1 and a2 be two complements of an element a L.
Then by definition of complement
a a1 I
a a1 0
...(3.5.1)
a a2 I
a a2 0
...(3.5.2)
3–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Consider
a1 = a1 0
= a1 (a a2)
[from (3.5.2)]
= (a1 a) (a1 a2)
[Distributive property]
= (a a1) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (3.5.1)]
= a1 a2
...(3.5.3)
Now Consider
a2 = a2 0
= a2 (a a1)
[from (3.5.2)]
= (a2 a) (a2 a1)
[Distributive property]
= (a a2) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (3.5.1)]
= a1 a2
...(3.5.4)
Hence, from (3.5.3) and (3.5.4),
a1 = a2
So, for bounded distributive lattice complement is unique.
Hasse diagram of [P(a, b, c), ] or [P(a, b, c), ] is shown in Fig. 3.5.1.
{a, b, c}
{a, b}
{a, c}
{a}
{b}
{b, c}
{c}
Fig. 3.5.1.
Greatest element is {a, b, c} and maximal element is {a, b, c}..
The least element is and minimal element is .
Que 3.6.
The directed graph G for a relation R on set A = {1, 2, 3,
4} is shown below :
1
2
3
4
Fig. 3.6.1.
3–6 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
i.
Verify that (A, R) is a poset and find its Hasse diagram.
ii. Is this a lattice ?
iii. How many more edges are needed in the Fig. 3.6.1 to extend
(A, R) to a total order ?
iv. What are the maximal and minimal elements ?
AKTU 2014-15, Marks 10
Answer
i.
The relation R corresponding to the given directed graph is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 1), (3, 4), (1, 4), (3, 2)}
R is a partial order relation if it is reflexive, antisymmetric and transitive.
Reflexive : Since aRa, a A . Hence, it is reflexive.
Antisymmetric : Since aRb and bRa then we get a = b otherwise aRb
or bRa.
Hence, it is antisymmetric.
Transitive: For every aRb and bRc we get aRc. Hence, it is transitive.
Therefore, we can say that (A, R) is poset. Its Hasse diagram is :
1
2
4
3
ii.
Fig. 3.6.2.
Since there is no lub of 1 and 2 and same for 2 and 4. The given poset is
not a lattice.
1 2 3 4
1 1 1 1
2 2 2
iii.
3 1 2 3 1
4 1 1 4
Only one edge (4, 2) is included to make it total order.
iv.
Maximals are {1, 2} and minimals are {3, 4}.
Que 3.7.
In a lattice if a b c, then show that
a.
ab=bc
b.
(a b) (b c) = (a b) (a c) = b
Answer
a.
Given : a b c
AKTU 2016-17, Marks 10
3–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Now
a b = least upper bound of a, b
= least {all upper bounds of a, b}
= least {b, c, ...}
[using a b c]
=b
and
...(3.7.1)
b c = greatest lower bound of b, c
= maximum {all lower bounds of b, c}
= maximum {b, a, ...}
=b
[using a b c]
...(3.7.2)
Eq. (3.7.1) and (3.7.2) gives, a b = b c
b.
(a b) (b c) (a b) (a c) = b
Consider, (a b) (b c)
= b b [using a b c and definition of and ]
=b
and
...(3.7.3)
(a b) (a c) = b c
=b
...(3.7.4)
From eq. (3.7.3) and (3.7.4), (a b) (b c) = (a b) (a c) = b.
Que 3.8.
a.
b.
Prove that every finite subset of a lattice has an LUB and a
GLB.
Give an example of a lattice which is a modular but not a
distributive.
AKTU 2016-17, Marks 10
Answer
a.
1.
The theorem is true if the subset has 1 element, the element being
its own glb and lub.
2. It is also true if the subset has 2 elements.
3. Suppose the theorem holds for all subsets containing 1, 2, ..., k
elements, so that a subset a1, a2, ..., ak of L has a glb and a lub.
4. If L contains more than k elements, consider the subset
{a1, a2, ..., ak + 1} of L.
5. Let w = lub (a1, a2, ..., ak).
6. Let t = lub (w, ak + 1).
7. If s is any upper bound of a1, a2, ..., ak + 1, then s is each of a1, a2, ...,
ak and therefore s w.
8. Also, s ak + 1 and therefore s is an upper bound of w and ak + 1.
9. Hence s t.
10. That is, since t each a1, t is the lub of a1, a2, ..., ak + 1.
11. The theorem follows for the lub by finite induction.
3–8 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
12. If L is finite and contains m elements, the induction process stops
when k + 1 = m.
b.
1.
2.
3.
4.
5.
The diamond is modular, but not distributive.
Obviously the pentagon cannot be embedded in it.
The diamond is not distributive :
y (x z) = y (y x) (y z) = 1
The distributive lattices are closed under sublattices and every
sublattice of a distributive lattice is itself a distributive lattice.
If the diamond can be embedded in a lattice, then that lattice has a
non-distributive sublattice, hence it is not distributive.
Que 3.9.
Show that the inclusion relation is a partial ordering
on the power set of a set S. Draw the Hasse diagram for inclusion on
the set P(S), where S = {a, b, c, d}. Also determine whether (P(S), ) is
AKTU 2018-19, Marks 07
a lattice.
Answer
Show that the inclusion relation () is a partial ordering on the power set of
a set S.
Reflexivity : A A whenever A is a subset of S.
Antisymmetry : If A and B are positive integers with A B and B A, then
A = B.
Transitivity : If A B and B C, then A C.
Hasse diagram :
{a, b, c, d}
{a, b, c}
{a, b}
{a, c}
{a}
{a, b, d}
{a, c}
{a, c, d}
{b, c, d}
{b, c}
{b, d}
{c}
{d}
{b}
Fig. 3.9.1.
{c, d}
3–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
(P(S), ) is not a lattice because ({a, b},{b, d}) has no lub and glb.
Que 3.10. Explain modular lattice, distributive lattice and bounded
AKTU 2017-18, Marks 10
lattice with example and diagram.
Answer
Modular distributive and bounded lattice : Refer Q. 3.2, Page 3–2C,
Unit-3.
Example :
Let consider a Hasse diagram :
1
a
c
b
0
Fig. 3.10.1.
Modular lattice :
0 a i.e., taking b = 0
b (a c) = 0 0 = 0, a (b c) = a c = 0
Distributive lattice :
For a set S, the lattice P(S) is distributive, since union and intersection each
satisfy the distributive property.
Bounded lattice : Since, the given lattice has 1 as greatest and 0 as least
element so it is bounded lattice.
Que 3.11.
Draw the Hasse diagram of (A, ), where
A = {3, 4, 12, 24, 48, 72} and relation be such that a b if a divides b.
AKTU 2017-18, Marks 07
Answer
Hasse diagram of (A, ) where A = {3, 4, 12, 24, 48, 72}
1
a
b
c
0
Fig. 3.11.1.
Que 3.12. For any positive integer D36, then find whether (D36, ‘|’)
is lattice or not ?
AKTU 2017-18, Marks 07
3–10 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
Answer
D36 = Divisor of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hasse diagram :
(1 3) = {3, 6}, (1 2) = {2, 4}, (2 6) = {6, 18}, (9 4) = {}
36
12
12
6
4
9
3
2
1
Since,
9 4 = {}
So, D36 is not a lattice.
PART-2
Boolean Algebra : Introduction, Axioms and Theorems of
Boolean Algebra.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 3.13.
What is Boolean algebra ? Write the axioms of Boolean
algebra. Also, describe the theorems of it.
Answer
A Boolean algebra is generally denoted by (B, +, ., 0, 1) where (B, +,.) is a
lattice with binary operations ‘+’ and ‘.’ called the join and meet respectively
and () is unary operation in B. The elements 0 and 1 are zero (least) and unit
(greatest) elements of lattice (B, +, .). B is called a Boolean algebra if the
following axioms are satisfied for all a, b, c in B.
Axioms of Boolean algebra :
If a, b, c B, then
1.
Commutative laws :
a.
2.
a.
3.
a+b=b+a
b.
a.b = b.a
b.
a.(b + c) = (a.b) = (a.c)
b.
a.1 = a
Distributive laws :
a + (b.c) = (a + b).(a + c)
Identity laws :
a.
a+0=a
3–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
4.
Complement laws :
a.
a + a = 1
b.
a.a = 0
b.
a.a = a
Basic theorems :
Let a, b, c B, then
1.
Idempotent laws :
a.
2.
a.
3.
b.
a.0 = 0
a + (a.b) = a
b.
a.(a + b) = a
b.
(a.b).c = a.(b.c)
b.
1 = 0
b.
(a.b) = a + b
Associative laws :
a.
5.
a+1=1
Absorption laws :
a.
4.
a+a=a
Boundedness (Dominance) laws :
(a + b) + c = a + (b + c)
Uniqueness of complement :
a + x = 1 and a.x = 0, then x = a
6.
Involution law : (a) = a
7.
a.
8.
De-Morgan’s laws :
a.
0 = 1
(a + b) = a.b
Que 3.14.
a.
b.
c.
d.
Prove the following theorems :
Absorption law : Prove that a, b, B
i.
a.(a + b) = a
ii. a + a.b = a
Idempotent law : Prove that a B, a + a = a and a.a = a.
De Morgan’s law : Prove that a, b, B
i.
(a + b) = a.b
ii. (a. b) = a + b
Prove that 0 = 1 and 1 = 0.
Answer
a.
i.
ii.
Absorption law :
To prove : a.(a + b) = a
Let
a.(a + b) =
=
=
=
=
To prove : a + a.b = a
Let
a + a.b =
=
=
=
=
(a + 0).(a + b)
a + 0.b
a + b.0
a+0
a
by Identity law
by Distributive law
by Commutative law
by Dominance law
by Identity law
a.1 + a.b
a.(1 + b)
a.(b + 1)
a.1
a
by Identity law
by Distributive law
by Commutative law
by Dominance law
by Identity law
3–12 C (CS/IT-Sem-3)
b.
c.
i.
ii.
d.
Lattices and Boolean Algebra
Idempotent law :
To prove : a + a = a and a.a = a
Let
a= a+0
by Identity law
= a + a. a
by Complement law
= (a + a).(a + a)
by Distributive law
= (a + a).1
by Complement law
= a+a
by Identity law
Now let
a = a.1
by Identity law
= a.(a + a)
by Complement law
= a.a + a.a
by Distributive law
= a.a + 0
by Complement law
= a.a
by Identity law
De Morgan’s law :
To prove : (a + b) = a.b
To prove the theorem we will show that
(a + b) + a.b = 1
Consider (a + b) + a.b = {(a + b) + a}.{(a + b) + b}by Distributive law
= {(b + a) + a}.{(a + b) + b}
by Commutative law
= {b + (a + a)}.{a + (b + b)}
by Associative law
= (b + 1).(a + 1)
by Complement law
= 1.1
by Dominance law
=1
...(3.14.1)
Also consider (a + b).ab = ab.(a + b)
by Commutative law
= ab.a + ab.b
by Distributive law
= a.(ab) + a.(b b)
by Commutative law
= (a. a).b + a(b. b)
by Associative law
= 0. b + a.0
by Complement law
= b.0 + a.0
by Commutative law
= 0+0
by Dominance law
=0
...(3.14.2)
From eq. (3.14.1) and (3.14.2), we get,
ab is complement of (a + b) i.e.(a + b) = a b.
To prove : (a. b) = a + b
Follows from principle of duality, that is, interchange operations + and •
and interchange the elements 0 and 1.
To prove : 0 = 1 and 1 = 0.
0 = (a a)
by Complement law
= a + (a)
by De Morgan’s law
= a + a
by Involution law
= a + a
by Commutative law
=1
by Complement law
Now
(0) = 1
0 = 1
1 = 0.
Discrete Structures & Theory of Logic
3–13 C (CS/IT-Sem-3)
Que 3.15. Define Boolean algebra. Show that in a Boolean algebra
meet and join operations are distributive to each other.
Answer
Boolean algebra : Refer Q. 3.10, Page 3–10C, Unit-3.
Meet and join operations are distributive :
1.
Let L be a poset under an ordering . Let a, b L.
2.
We define :
a b (read “a join b”) as the least upper bound of a and b, and
a b(read “a meet b”) as greatest lower bound of a and b.
3.
Since the join and meet operation produce a unique result in all cases
where they exist, we can consider them as binary operations on a set if
they always exist.
4.
A lattice is a poset L (under ) in which every pair of elements has a lub
and a glb.
5.
Since a lattice L is an algebraic system with binary operations and , it
is denoted by [L, , ].
6.
Let us consider,
a. [P(A), , ] is a lattice for any set A and
b. The join operation is the set operation of union and the meet
operation is the operation of intersection; that is, = and = .
7.
It can be shown that the commutative laws, associative laws, idempotent
laws, and absorption laws are all true for any lattice.
8.
An example of this is clearly [P(A); , ], since these laws hold in the
algebra of sets.
9.
This lattice is also distributive such that join is distributive over meet
and meet is distributive over join.
PART-3
Algebraic Manipulation of Boolean Expressions, Simplification
of Boolean Functions, Karnaugh Maps.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 3.16. Express each Boolean expression as SOP and then in its
complete sum-of-products form.
3–14 C (CS/IT-Sem-3)
a.
b.
Lattices and Boolean Algebra
E = x(xy + xy + yz)
E = z(x+ y) + y.
Answer
a.
b.
E = x.x.y + x.x.y + x.y.z = x.y + x.y.z = x.y'
( x.x.y = 0 and x.y is contained in x.y.z)
Complete SOP form = x.y (z + z)
( z is missing)
= x.y.z + x.y. z
E = z.(x+ y) + y = x.z + y.z + y
Then
E = x.z(y + y) + y.z.(x + x) + y.(x + x) (z + z)
(by Complement law)
= x.y.z + x.y.z + x.y.z + x.y.z + x.y.z + x.z.y
+ x.y.z+ x.y.z
= x.y.z + x.y.z + x.y.z + x.y.z + x.y.z + x.y.z
Que 3.17.
Define a Boolean function of degree n. Simplify the
following Boolean expression using Karnaugh maps
xyz + xy'z + x'y'z + x'yz + x'y'z'
Answer
Boolean function of degree n :
1.
Let B = {0, 1}. Then Bn = {(x1, x2 ...., xn)|xi B for 1 i n} is the set of all
possible n-tuples of 0s and 1s.
2.
The variable x is called a Boolean variable if it assumes values only
from B, that is, if its only possible values are 0 and 1.
3.
A function from Bn to B is called a Boolean function of degree n.
4.
For example, the function F(x, y) = xy from the set of ordered pairs of
Boolean variables to the set {0, 1}, is a Boolean function of degree z
with F(1, 1) = 1, F(1, 0) = 0, F(0, 1) = 0 and F(0, 0) = 0.
Numerical : The Karnaugh map for the given function is :
xyz + xyz + xyz + xyz + xyz
yz
x
x
x
yz
yz
yz
1
1
1
1
1
yz
Fig. 3.17.1.
Then the simplified expression is : z + x'y'.
Que 3.18.
Simplify the following Boolean functions using three
variable maps :
3–15 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
a.
b.
F (x, y, z) = (0, 1, 5, 7)
F (x, y, z) = (1, 2, 3, 6, 7)
Answer
a.
f(x, y, z) = (0, 1, 5, 7)
z
0
xy
00 0
1
1
01 2
3
11 6
7
10 4
5
f=xy+xz
Fig. 3.18.1.
b.
f (x, y, z) = (1, 2, 3, 6, 7)
yz
x
0
1
00
0
4
01
1
5
11
3
7
10
2
6
Fig. 3.18.2.
f= xz+y
Que 3.19.
Simplify the following Boolean function using K-map :
F(x, y, z) = (0, 2, 3, 7)
AKTU 2017-18, Marks 07
Answer
x
yz
0
00
01
1
1
11
10
1
1
1
F = x z yz
Que 3.20.
Simplify the following Boolean expressions using
K-map :
a. Y = ((AB) + A + AB)
b. A B C D + A B C D + A B CD + A B CD = A B
AKTU 2015-16, Marks 10
Answer
a.
Y = ((AB) + A + AB)
= ((AB)) (A + (AB))
= (AB) ((A) (AB))
3–16 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
= AB (A (A + B))
= AB (AA + AB)
= AB(0 + AB) = AB AB
= ABB
=0
Here, we find that the expression is not in minterm. For getting minterm,
we simplify and find that its value is already zero. Hence, no need to use
K-map for further simplification.
b.
A B C D + A B C D + A B CD + A B CD = A B
= A BCD + ABCD + ABCD + ABCD
CD AB
A B C D
1
A B C D
1
AB CD
1
AB CD
1
Fig. 3.20.1.
On simplification by K-map, we get AB corresponding to all the four
one’s.
Que 3.21.
Find the Boolean algebra expression for the following
AKTU 2016-17, Marks 7.5
system.
A
B
C
Fig. 3.21.1.
Answer
A
B
C
ABC
ABC + A(B + C )
A.(B + C )
B
B + C
C
Fig. 3.21.2.
3–17 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Que 3.22.
Find the Sum-Of-Products and Product-Of-s um
expansion of the Boolean function F(x, y, z) = (x + y) z.
AKTU 2018-19, Marks 07
Answer
F(x, y, z) = (x + y)z
x
y
z
x+y
z
(x + y)z
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
0
0
1
0
0
1
1
1
0
1
1
1
0
0
0
1
0
1
1
1
0
0
1
0
0
0
0
0
0
0
1
0
Sum-Of-Product :
F(x, y, z) = xyz + xyz + xyz
Product-Of-Sum :
F(x, y, z) = (x + y + z)(x + y+ z)(x+ y + z)(x+ y+ z)
(x+ y+ z')
PART-4
Logic Gates, Digital Circuits and Boolean Algebra.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 3.23.
a.
b.
Consider the Boolean function.
f(x1, x2, x3, x4) = x1 + (x2. (x1 + x4) + x3. (x2 + x4))
i.
Simplify f algebraically
ii.
Draw the logic circuit of the f and the reduction of the f.
Write the expres sions E 1 = (x + x * y) + (x/y) and
E2 = x + ((x * y + y)/y), into
3–18 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
i.
Prefix notation
ii.
Postfix notation
AKTU 2014-15, Marks 10
Answer
a. i. f(x1, x2, x3, x4) = x1 + (x2.(x1 + x4) + x3.(x2+ x4)
= x1 + x2.x1 + x2.x4 + x3.x2 + x3.x4
= x1 + x2 + x2.x4 + x3.x2 + x3.x4
= x1 + x2.(1 + x4) + x3.x2 + x3.x4
= x1 + x2 + x3.x2 + x3.x4
= x1 + x2 + x3 + x3.x4
= x1 + x2 + x3.(1 + x4)
= x1 + x2 + x3
ii.
Logic circuit :
x1
x2
x3
x1 + x2 + x3
Fig. 3.23.1.
Reduction of f :
f (x1, x2, x3, x4) = x1 + (x2.(x1 + x4) + x3.(x2 + x4)
= x1 + (x2.x1 + x2.x4) + (x3.x2 + x3.x4)
= x1 + x2.x1 + x2.x4 + x3.x2 + x3.x4
b.
E1 = (x + x * y) + (x/y)
Binary tree is :
+
+
x
/
x
*
x
y
Fig. 3.23.2.
Prefix : ++ x * x y / xy
Postfix : xx y * + xy / +
E2 = x + ((x * y + y)/y)
y
3–19 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Binary tree is :
+
/
x
y
+
y
*
y
x
Fig. 3.23.3.
Prefix : + x / + * x y y y
Postfix : x x y * y + y / +
Que 3.24.
Construct circuits that produce the following output :
F(X, Y, Z) = (X + Y + Z) ( XYZ )
Answer
The logic network is shown below :
X
Y
Z
X+Y+Z
(X + Y + Z) X Y Z
X
Y
X
Y
XYZ
Z
Z
Fig. 3.24.1.
Que 3.25.
Draw the logic network corresponding to the following
Boolean expressions :
i.
ii.
xy + x y
XYZ + XYZ + XY
3–20 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
Answer
i.
The logic network is shown in Fig. 3.25.1.
x
y
xy + xy
x
y
xy
y
Fig. 3.25.1.
ii.
The logic network shown in Fig. 3.25.2.
X
Y
Z
X
XY Z
Y
Z
XY Z + XYZ + XY
X YZ
XY
Fig. 3.25.2.
VERY IMPORTANT QUESTIONS
Following questions are very important. These questions
may be asked in your SESSIONALS as well as
UNIVERSITY EXAMINATION.
Q. 1. If the lattice is represented by the Hasse diagram given
below :
i. Find all the complements of ‘e’.
ii. Prove that the given lattice is bounded complemented
lattice.
b
a
c
f
d
e
Fig. 1.
3–21 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Ans. Refer Q. 3.3.
Q. 2. Define a lattice. For any a, b, c, d in a lattice (A, ) if a b and
c d then show that a c b d and a c b d.
Ans. Refer Q. 3.4.
Q. 3. Let L be a bounded distributed lattice, prove if a complement
exists, it is unique. Is D12 a complemented lattice ? Draw
the Hasse diagram of [P (a, b, c), <], (Note : ‘<’ stands for
subset). Find greatest element, least element, minimal
element and maximal element.
Ans. Refer Q. 3.5.
Q. 4. The directed graph G for a relation R on set A = {1, 2, 3, 4} is
shown below :
1
2
3
4
Fig. 2.
i. Verify that (A, R) is a poset and find its Hasse diagram.
ii. Is this a lattice ?
iii. How many more edges are needed in the Fig. 2 to extend
(A, R) to a total order ?
iv. What are the maximal and minimal elements ?
Ans. Refer Q. 3.6.
Q. 5. In a lattice if a b c, then show that
a. a b = b c
b. (a b) (b c) = (a b) (a c) = b
Ans. Refer Q. 3.7.
Q. 6.
a. Prove that every finite subset of a lattice has an LUB and a
GLB.
b. Give an example of a lattice which is a modular but not a
distributive.
Ans. Refer Q. 3.8.
Q. 7. Show that the inclusion relation is a partial ordering on
the power set of a set S. Draw the Hasse diagram for
inclusion on the set P(S), where S = {a, b, c, d}. Also determine
whether (P(S), ) is a lattice.
Ans. Refer Q. 3.9.
3–22 C (CS/IT-Sem-3)
Lattices and Boolean Algebra
Q. 8. Explain modular lattice, distributive lattice and bounded
lattice with example and diagram.
Ans. Refer Q. 3.10.
Q. 9. Draw the Hasse diagram of (A, ), where A = {3, 4, 12, 24, 48,
72} and relation be such that a b if a divides b.
Ans. Refer Q. 3.11.
Q. 10. For any positive integer D36, then find whether (D36, ‘|’) is
lattice or not ?
Ans. Refer Q. 3.12.
Q. 11. Simplify the following Boolean function using K-map :
F(x, y, z) = (0, 2, 3, 7)
Ans. Refer Q. 3.19.
Q. 12. Simplify the following Boolean expres sions us ing
K-map :
a. Y = ((AB) + A + AB)
b. A B C D + A B C D + A B CD + A B CD = A B
Ans. Refer Q. 3.20.
Q. 13. Find the Boolean algebra expression for the following
system.
A
B
C
Fig. 3.
Ans. Refer Q. 3.21.
Q. 14. Find the Sum-Of-Products and Product-Of-sum expansion
of the Boolean function F(x, y, z) = (x + y) z.
Ans. Refer Q. 3.22.
Q. 15. Consider the Boolean function.
a. f(x1, x2, x3, x4) = x1 + (x2. (x1 + x4) + x3. (x2 + x4))
i. Simplify f algebraically
ii. Draw the logic circuit of the f and the reduction of the
f.
b. Write the expres sions E 1 = (x + x * y) + (x/y) and
E2 = x + ((x * y + y)/y), into
i. Prefix notation
ii. Postfix notation
Ans. Refer Q. 3.23.
Discrete Structures & Theory of Logic
4
4–1 C (CS/IT-Sem-3)
Propositional Logic
and Predicate Logic
CONTENTS
Part-1
:
Propositional Logic : .................................. 4–2C to 4–6C
Proposition, Well Formed
Formula, Truth Tables
Part-2
:
Tautology, Satisfiability, ......................... 4–6C to 4–14C
Contradiction, Algebra of
Proposition, Theory of Inference
Part-3
:
Predicate Logic : ..................................... 4–15C to 4–19C
First Order Predicate,
Well Formed Formula
of Predicate, Quantifiers,
Inference Theory of
Predicate Logic
4–2 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
PART-1
Propositional Logic : Proposition, Well Formed Formula,
Truth Tables.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 4.1.
Define the term proposition. Also, explain compound
proposition with example.
Answer
Proposition : Proposition is a statement which is either true or false but not
both. It is a declarative statement. It is usually denoted by lower case letters
p, q, r, s, t etc. They are called Boolean variable or logic variable.
For example :
1.
Dr. A.P.J. Abdul Kalam was Prime Minister of India.
2.
Roses are red.
3.
Delhi is in India
(1) proposition is false whereas (2) and (3) are true.
Compound proposition : A compound proposition is formed by composition
of two or more propositions called components or sub-propositions.
For example :
1.
Risabh is intelligent and he studies hard.
2.
Sky is blue and clouds are white.
Here first statement contain two propositions ‘‘Risabh is intelligent’’ and ‘‘he
studies hard’’ whereas second statement contain propositions ‘‘sky is blue’’
and ‘‘clouds are white’’. As both statements are formed using two propositions.
So they are compound propositions.
Que 4.2.
Discuss connectives in detail with truth tables.
Answer
1.
The words or phrases used to form compound proposition are called
connectives.
2.
There are five basic connectives as shown in the Table 4.2.1.
4–3 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Table. 4.2.1.
S. No.
Connective words
Name of connective
Symbol
1.
Not
Negation
~ or or –
2.
And
Conjunction
3.
Or
Disjunction
4.
If-then
Implication
5.
If and only if
Biconditional
i.
Negation : If P is a proposition then negation of P is a proposition
which is true when p is false and false when p is true. It is denoted
by ~ p or or p’ or p .
Truth table :
ii.
p
~p
T
F
F
T
Conjunction : If p and q are two propositions then conjunction of
p and q is a proposition which is true when both p and q are true
otherwise false. It is denoted by p q.
Truth table :
p
q
pq
T
T
T
T
F
F
F
T
F
F
F
F
iii. Disjunction : If p and q be two propositions, then disjunction of p
and q is a proposition which is true when either one of p or q or both
are true and is false when both p and q are false and it is denoted by
p q.
Truth table :
p
q
pq
T
T
T
T
F
T
F
T
T
F
F
F
4–4 C (CS/IT-Sem-3)
Que 4.3.
Propositional Logic & Predicate Logic
What do you mean by well formed formula ?
Answer
1.
Well formed formula is defined as a mathematical expression
represented in well defined form and uses parenthesis, braces and
square brackets to avoid the ambiguity.
2.
Logical statements are also represented in well defined form using
parenthesis according to priority of operations. To generate well formed
formula recursively, following rules are used :
i.
An atomic statement P is a well formed formula.
ii.
If P is well formed formula then ~ P is also a well formed.
iii.
If P and Q are well formed formulae then (P Q), (P Q), (P Q)
and (P Q) are also well formed formulae.
iv. A statement consists of variables, parenthesis and connectives is
recursively a well formed formula iff it can be obtained by applying
the above three rules.
For example :
1. (P (P Q)
2. ((P Q) R)
Que 4.4.
i.
Write short notes on :
Truth table
ii.
Logical equivalence
Answer
i.
ii.
Truth table : A truth table is a table that shows the truth value of a
compound proposition for all possible cases.
p
q
(p q)
p
q
(p q)
p
~p
T
T
T
T
T
T
T
F
T
F
F
T
F
T
F
T
F
T
F
F
T
T
F
F
F
F
F
F
Logical equivalence : If two propositions P (p, q, .....) and Q (p, q, ...)
where p, q, ..... are propositional variables, have the same truth values
in every possible case, the propositions are called logically equivalent or
simply equivalent, and denoted by
P (p, q, .......) Q (p, q, ........)
Que 4.5.
Explain the following terms with suitable example :
i.
Conjunction
ii. Disjunction
iii. Conditional
4–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
iv. Converse
v.
AKTU 2014-15, Marks 10
Contrapositive
OR
Define inverse.
Answer
i.
Conjunction : If p and q are two statements, then conjunction of p and
q is the compound statement denoted by p q and read as
“p and q”. Its truth table is,
p
q
pq
T
T
T
T
F
F
F
T
F
F
F
F
Example :
p : Ram is healthy.
q : He has blue eyes.
p q : Ram is healthy and he has blue eyes.
ii.
Disjunction : If p and q are two statements, the disjunction of
p and q is the compound statement denoted by p q and it is read as “p
or q”. Its truth table is,
p
q
pq
T
T
T
T
F
T
F
T
T
F
F
F
Example :
p : Ram will go to Delhi.
q : Ram will go to Calcutta.
p q : Ram will go to Delhi or Calcutta.
iii. Conditional : If p and q are propositions. The compound proposition if
p then q denoted by p q or p q and is called conditional proposition
or implication. It is read as “If p then q” and its truth table is,
4–6 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
p
q
pq
T
T
T
T
F
F
F
T
T
F
F
T
Example :
p : Ram works hard.
q : He will get good marks.
p q : If Ram works hard then he will get good marks.
For converse and contrapositive :
Let
p : It rains.
q : The crops will grow.
iv. Converse : If p q is an implication then its converse is given by q p
states that S : If the crops grow, then there has been rain.
v.
Contrapositive : If p q is an implication then its contrapositive is
given by q p states that,
t : If the crops do not grow then there has been no rain.
Inverse :
If p q is implication the inverse of p q is ~ p ~ q.
Consider the statement
p : It rains.
q : The crops will grow
The implication p q states that,
r : If it rains then the crops will grow.
The inverse of the implication p q, namely ~ p ~ q states that.
u : If it does not rain then the crops will not grow.
PART-2
Tautology, Satisfiability, Contradiction, Algebra of Proposition,
Theory of Inference.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Discrete Structures & Theory of Logic
Que 4.6.
4–7 C (CS/IT-Sem-3)
Explain tautologies, contradictions, satisfiability and
contingency.
Answer
1.
Tautology : Tautology is defined as a compound proposition that is
always true for all possible truth values of its propositional variables and
it contains T in last column of its truth table.
Propositions like,
i.
The doctor is either male or female.
ii.
Either it is raining or not.
are always true and are tautologies.
2.
Contradiction : Contradiction is defined as a compound proposition
that is always false for all possible truth values of its propositional
variables and it contains F in last column of its truth table.
Propositions like,
i.
x is even and x is odd number.
ii.
Tom is good boy and Tom is bad boy.
are always false and are contradiction.
3.
Contingency : A proposition which is neither tautology nor
contradiction is called contingency.
Here the last column of truth table contains both T and F.
4.
Satisfiability :
A compound statement formula A (P1, P2, ... Pn) is said to be satisfiable,
if it has the truth value T for at least one combination of truth value of
P1, P2,.... Pn.
Que 4.7.
Write short note on algebra of propositions.
Answer
Proposition satisfies various laws which are useful in simplifying complex
expressions. These laws are listed as :
1.
Idempotent laws :
a. p p p
b. p p p
2.
Associative laws :
a. (p q) r p (q r)
b. (p q) r p (q r)
3.
Commutative laws :
a. p q q p
b. p q q p
4–8 C (CS/IT-Sem-3)
4.
Distributive laws :
a.
p (q r) (p q) (p r)
b.
p (q r) (p q) (p r)
5.
Identity laws :
a. p F p
b. p T T
c. p F F
d. p T p
6.
Complement laws :
7.
a.
p ~P T
b.
p ~p F
c.
~T F
d.
~F T
Involution law :
a.
8.
9.
Propositional Logic & Predicate Logic
~(~p) p
De Morgan's laws :
a.
~ (p q) ~p ~q
b.
~ (p q) ~p ~q
Absorption laws :
a.
p (p q) p
b
p (p q) p
These laws can easily be verified using truth table.
Que 4.8.
Discuss theory of inference in propositional logic.
Answer
Rules of inference are the laws of logic which are used to reach the given
conclusion without using truth table. Any conclusion which can be derived
using these laws is called valid conclusion and hence the given argument is
valid argument.
1.
Modus ponens (Law of detachment) : By this rule if an implication
p q is true and the premise p is true then we can always conclude that
q is also true.
The argument is of the form :
p q
p
2.
q
Modus tollens (Law of contraposition) : By this rule if an implication
p q is true and conclusion q is false then the premise p must be false.
The argument is of the form :
Discrete Structures & Theory of Logic
4–9 C (CS/IT-Sem-3)
p q
~q
~p
3.
Hypothetical syllogism : By this rule whenever the two implications
p q and q r are true then the implication p r is also true.
The argument is of the form :
p q
q r
p r
4.
Disjunctive syllogism : By this rule if the premises p q and ~ q are
true then p is true.
The argument is of the form :
p q
~q
p
5.
Addition : By this rule if p is true then p q is true regardless the truth
value of q.
The argument is of the form :
p
6.
Simplification : By this rule if p q is true then p is true.
The argument is of form :
p q
p q
p or q
7.
Conjunction : By this rule if p and q are true then p q is true.
The argument is of the form :
p q
p
q
p q
8.
Constructive dilemma : By this rule if (p q) (r s) and p r are
true then q s is true.
The argument is of form :
( p q) ( r s)
p r
q s
9.
Destructive dilemma : By this rule if (p q) (r s) and ~ q s
are true.
The argument is of the form :
4–10 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
( p q) (r s)
~qs
pr
10. Absorption : By this rule if p q is true then p (p q) is true.
The argument is of the form :
p q
p ( p q)
Que 4.9.
What do you mean by valid argument ? Are the following
arguments valid ? If valid, construct a formal proof ; if not, explain
why.
For students to do well in discrete structure course, it is necessary
that they study hard. Students who do well in courses do not skip
classes. Student who study hard do well in courses. Therefore
students who do well in discrete structure course do not skip class.
Answer
Valid arguments :
1. An argument P1, P2, ..., Pn Q is said to be valid if Q is true whenever
all the premises P1, P2, ..., Pn are true.
2. For example : Consider the argument : p q, q
p.
C
P
P
p
q
p q
T
T
T
T
F
F
F
T
T
F
F
T
where P denotes the premise and C denotes the conclusion.
3.
From the truth table we can see in first and third rows both the premises
q and p q are true, but the conclusion p is false in third row. Therefore,
this is not a valid argument.
4.
First and third rows are called critical rows.
5.
This method to determine whether the conclusion logically follows
from the given premises by constructing the relevant truth table is
called truth table technique.
6.
Also, we can say the argument P1, P2, ..., Pn
Q is valid if and only if
the propo sitio n P 1 P 2 ... P n is true o r we can say if
P1 P2 ... Pn Q is a tautology.
For example : Consider the argument p q, p
q.
4–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Then from the truth table :
p
q
pq
T
T
T
T
T
T
F
F
F
T
F
T
T
F
T
F
F
T
F
T
p p q p (p q ) q
p (p q) q is a tautology since the last column contains T only.
p q, p
q is a valid argument.
Numerical :
Let the propositional variables be :
p Do well in the course.
q They study hard.
r Do not skip classes.
1. For students to do well in discrete structure course, it is necessary that
they study hard : p q
2. Students who do well in the courses do not skip classes : p r
3. Students who study hard do well in courses : q p
4. Therefore, students who do well in discrete structure course do not skip
classes : p r
Therefore, we have,
Given :
pq
pr
q p Conclusion : p r
I
II
III
IV
Proof : Taking III and II together we get
q p, p r gives q r
(Using hypothetical syllogism)
V
Now taking I and V
p q and q r we get p r
(Using hypothetical syllogism)
Hence, p r is conclusion, so it is valid.
Yes, the statement is valid.
Que 4.10.
i.
ii.
a.
b.
c.
d.
Show that ((p q) ~ (~ p (~ q ~ r))) (~ p ~q) (~p r) is a
tautology without using truth table.
Rewrite the following arguments using quantifiers, variables
and predicate symbols :
All birds can fly.
Some men are genius.
Some numbers are not rational.
There is a student who likes mathematics but not geography.
AKTU 2014-15, Marks 10
OR
Show that ((p q) ~ (~p (~q ~r))) (~p ~q) (~ p r) is a
tautology without using truth table.
AKTU 2018-19, Marks 07
4–12 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
Answer
i.
We have
((p q) ~ (~ p (~ q r))) (~ p ~ q) (~ p r)
((p q) ~ (~ p ~ (q r))) (~ (p q) ~ ( p r))
(Using De Morgan's Law)
[(p q)] (p (q r)) ~ ((p q) (p r))
[(p q) (p q) (p r)] ~ (( p q) (p r))
(Using Distributive Law)
[((p q) (p q)] (p r) ~ ((p q) (p r))
((p q) (p r)) ~ ((p q) (p r))
x ~ x where x = (p q) (p r)
T
a. x [B(x) F(x)]
b. x [M(x) G(x)]
c. ~[ (x) (N(x) R(x)]
d. x [S(x) M(x) ~ G(x)]
ii.
Que 4.11.
“If the labour market is perfect then the wages of all
persons in a particular employment will be equal. But it is always
the case that wages for such persons are not equal therefore the
labour market is not perfect”. Test the validity of this argument
AKTU 2014-15, Marks 10
using truth table.
Answer
Let
p1 : The labour market is perfect.
p2 : Wages of all persons in a particular employment will be equal.
~p2 : Wages for such persons are not equal.
~p1 : The labour market is not perfect.
The premises are p1 p2 , ~ p2 and the conclusion is ~ p1. The argument
p1 p2, ~ p2 ~ p1 is valid if ((p1 p2) ~ p2) ~ p1 is a tautology.
Its truth table is,
~p2 p1 p2 (p1 p2) ~p2 (p1p2 ~ p2) ~p1
p1
p2
~p1
T
T
F
F
T
F
T
T
F
F
T
F
F
T
F
T
T
F
T
F
T
F
F
T
T
T
T
T
Since ((p1 p2) ~p2) ~p1 is a tautology. Hence, this is valid argument.
Que 4.12.
What is a tautology, contradiction and contingency ?
Show that (p q) ( p r) (q r) is a tautology, contradiction or
contingency.
AKTU 2018-19, Marks 07
4–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Answer
Tautology, contradiction and contingency : Refer Q. 4.6, Page 4–7C,
Unit-4.
Proof : ((p q) (~ p r)) (q r)
p
q
r
~P
F
F
F
T
F
F
T
F
T
F
(p q) (~ p r) (A B)
=A
=B
=C
(q r) C D
=D
F
T
T
F
F
T
F
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
T
T
T
T
T
F
F
F
T
F
T
F
F
T
F
T
F
T
T
T
T
T
T
T
F
F
T
F
T
T
T
T
T
T
F
T
T
T
T
T
So, ((p q) (~ p r)) (q r) is contingency.
Que 4.13.
i.
ii.
Find a compound proposition involving the propositional
variables p, q, r and s that is true when exactly three
propositional variables are true and is false otherwise.
Show that the hypothesis “It is not sunny this afternoon and it
is colder than yesterday”, “We will go swimming only if it is
sunny”, “If we do not go swimming, then we will take a canoe
trip.” and “If we take a canoe trip, then we will be home by
sunset” lead to the conclusion “We will be home by sunset.”
OR
Show that the premises “It is not sunny this afternoon and it is
colder than yesterday,” “We will go swimming only if it is sunny,”
“If we do not go swimming, then we will take a canoe trip.” and
“If we take a canoe trip, then we will be home by sunset” lead to
the conclusion “We will be home by sunset.”
AKTU 2018-19, Marks 07
Answer
i.
ii.
The compound proposition will be : (p q r) s
Let p be the proposition “It is sunny this afternoon”, q be the proposition
“It is colder than yesterday”, r be the proposition “We will go swimming”,
s be the proposition ‘‘We will take a canoe trip’’, and t be the proposition
“We will be home by sunset”.
4–14 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
Then the hypothesis becomes p q, r p, r s, and s t. The
conclusion is simply t.
We construct an argument to show that our hypothesis lead to the
conclusion as follows :
S. No.
Step
Reason
1.
pq
Hypothesis
2.
p
Simplification using step 1
3.
rp
Hypothesis
4.
r
Modus tollens using steps 2 and 3
5.
rs
Hypothesis
6.
s
Modus ponens using steps 4 and 5
7.
st
Hypothesis
8.
t
Modus ponens using steps 6 and 7
Que 4.14.
Show that : (r ~ q, r S, S ~ q, p q) ~ p are
AKTU 2017-18, Marks 07
inconsistent.
Answer
Following the indirect method, we introduce p as an additional premise and show
that this additional premise leads to a contradiction.
{1}
(1) p q
Rule P
{2}
(2) p
Rule P (assumed)
{1, 2}
(3) q
Rule T, (1), (2) and modus ponens
{4}
(4) s q
Rule P
s
{1, 2, 4}
(5)
Rule T, (3), (4) and modus tollens
{6}
(6) r s
Rule P
{1, 2, 4, 6}
(7) r
Rule T, (5), (6) disjunctive syllogism
{8}
(8) r q
Rule P
r
q
{8}
(9)
Rule T, (8) and EQ16 (p q p q)
{8}
{1, 2, 4, 6}
(10) r q
(11) r q
{1, 2, 4, 6, 8} (12) r q r q
Rule T, (8) and De Morgan’s law
Rule T, (7), (3) and conjunction
Rule T, (10), (11) and conjunction.
Since, we know that set of formula is inconsistent if their conjunction implies
contradiction. Hence it leads to a contradiction. So, it is inconsistent.
Discrete Structures & Theory of Logic
4–15 C (CS/IT-Sem-3)
PART-3
Predicate Logic : First Order Predicate, Well Formed Formula of
Predicate, Quantifiers, Inference Theory of Predicate Logic.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 4.15.
1.
2.
Write a short note on
First order logic
Quantifiers
Answer
1.
2.
First order logic :
i.
First order logic is the extension of propositional logic by generalizing
and quantifying the propositions over given universe of discourse.
ii.
In first order logic every individual has the property p (say).
iii.
It is also called first order predicate calculus.
iv.
Predicate calculus is generalization of propositional calculus.
Predicate calculus allows us to manipulate statements about all or
something.
v.
Universe of Discourse (UD) : It is the set of all possible values that
can be substituted in place of predicate variable.
Quantifiers : There are following two types of quantifiers :
i.
Universal quantifier : Let p(x) be a propositional function defined
on set A. Consider the expression.
( x A) P(x) or x P(x)
...(4.15.1)
Here the symbol ‘‘’’ is read as ‘‘for all’’ or ‘‘for every’’ and is called
universal quantifier. Then the statement (4.15.1) is read as ‘‘For
every x A, P(x) is true.’’
ii.
Existential quantifier : Let Q(x) be a propositional function defined
on set B. Consider the expression
(x B) Q(x) or x Q(x)
...(4.15.2)
Here the symbol ‘‘’’ is read as ‘‘for some’’ or ‘‘for at least one’’ or
‘‘there exists’’ and is called existential quantifier.
Then the statement (4.15.2) is read as ‘‘For some x B, Q(x) is
true’’.
Que 4.16.
Explain rules of inference in predicate logic.
4–16 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
Answer
Rule of inference is a logical form consisting of function which takes premises,
analyzes their syntax and returns a conclusion.
Rules of inference :
i.
Universal specification : By this rule if the premise ( x) P (x) is true
then P(c) is true where c is particular member of UD.
(x) P ( x)
ii.
P ( c)
Universal generalization : By this rule if P(c) is true for all c in UD
then (x) P(x) is true.
P ( c)
(x) P ( x)
x is not free in any of given premises.
iii. Existential specification : By this rule if (x) P(x) is true then P(x) is
true for some particular member of UD.
(x) P ( x)
P (c)
c is some member of UD.
iv. Existential generalization : By this rule if P(c) is true for some
particular member c in UD, then ( x) P (x) is true
P ( c)
v.
(x) P ( x)
c is some member of UD.
Universal modus ponens : By this rule if P(x) Q(x) is true for every
x and P(c) is true for some particular member c in UD then Q(c) is true.
( x) P ( x) Q( x)
P (c)
Q(c)
vi. Universal modus tollens : By this rule if P(x) Q(x) is true for every
x and ~ Q(c) is true for some particular c in UD then ~ Q(c) is true.
( x) P ( x) Q( x)
~ Q(c)
~ P (c)
Que 4.17.
Write the symbolic form and negate the following
statements :
a. Everyone who is healthy can do all kinds of work.
b. Some people are not admired by everyone.
c. Everyone should help his neighbours, or his neighbours will
not help him.
AKTU 2016-17, Marks 10
4–17 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Answer
a.
Symbolic form :
Let P(x): x is healthy and Q(x): x do all work
x(P(x) Q(x))
Negation : ( x (P(x) Q(x))
b.
Symbolic form :
Let P(x) : x is a person
A(x, y) : x admires y
The given statement can be written as “There is a person who is not
admired by some person” and it is (x) ( y)[P(x) P(y) A(x, y)]
Negation : ( x) ( y) [P (x) P(y) A(x, y)]
c.
Symbolic form :
Let N(x, y) : x and y are neighbours
H(x, y) : x should help y
P(x, y) : x will help y
The statement can be written as “For every person x and every person
y, if x and y are neighbours, then either x should help y or y will not help
x” and it is ( x) ( y)[N(x, y) (H(x, y) P(y, x))]
Negation : ( x) ( y) [N(x, y) (H (x, y) P (y, x))]
Que 4.18. Express the following statements using quantifiers and
logical connectives.
a. Mathematics book that is published in India has a blue cover.
b. All animals are mortal. All human being are animal. Therefore,
all human being are mortal.
c. There exists a mathematics book with a cover that is not blue.
d. He eats crackers only if he drinks milk.
e. There are mathematics books that are published outside India.
f.
Not all books have bibliographies.
AKTU 2015-16, Marks 10
Answer
a.
P(x) : x is a mathematic book published in India
Q(x) : x is a mathematic book of blue cover
x P(x) Q(x).
b.
P(x) : x is an animal
Q(x) : x is mortal
x P(x) Q(x)
4–18 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
R(x) : x is a human being
x R(x) P(x).
c.
P(x) : x is a mathematics book
Q(x) : x is not a blue color
x, P(x) Q(x).
d.
P(x) : x drinks milk
Q(x) : x eats crackers
for x, if P(x) then Q(x).
or x, P(x) Q(x).
e.
P(x) : x is a mathematics book
Q(x) : x is published outside India
x P(x) Q(x).
f.
P(x) : x is a book having bibliography ~ x, P(x).
Que 4.19.
i.
Express this statement using quantifiers : “Every student in
this class has taken some course in every department in the
school of mathematical sciences”.
ii.
If x yP( x , y) is true, does it necessarily follow that yP( x , y) is
true ? Justify your answer.
Answer
i.
x p(x) z Q(y, z)
where P(x) is student of class.
Q(y, z) is the course from department.
ii.
Let P(x, y) be x + y = 3 and x, y belong to some set of integers xyP ( x, y) is
true means for all x there exists some y for which x + y = 3 is true but for
all y we conclude that P(x, y) will not be true.
Que 4.20.
Trans late the following sentences in quantified
expressions of predicate logic.
i.
All students need financial aid.
ii. Some cows are not white.
iii. Suresh will get if division if and only if he gets first div.
iv. If water is hot, then Shyam will swim in pool.
v. All integers are either even or odd integer.
AKTU 2017-18, Marks 07
Discrete Structures & Theory of Logic
4–19 C (CS/IT-Sem-3)
Answer
i.
x [S(x) F(x)]
ii.
~ [(x) (C(x) W(x))]
iii.
Sentence is incorrect so cannot be translated into quantified expression.
iv.
W(x) : x is water
H(x) : x is hot
S(x) : x is Shyam
P(x) : x will swim in pool
x [((W(x) H(x)) (S(x) P(x))]
v.
E(x) : x is even
O(x) : x is odd
x (E(x) O(x))
VERY IMPORTANT QUESTIONS
Following questions are very important. These questions
may be asked in your SESSIONALS as well as
UNIVERSITY EXAMINATION.
Q. 1.
i.
ii.
iii.
iv.
v.
Explain the following terms with suitable example :
Conjunction
Disjunction
Conditional
Converse
Contrapositive
Ans. Refer Q. 4.5.
Q. 2. Show that ((p q) ~ (~p (~q ~r))) (~p ~q) (~ p r) is
a tautology without using truth table.
Ans. Refer Q. 4.10.
Q. 3. “If the labour market is perfect then the wages of all persons
in a particular employment will be equal. But it is always
the case that wages for such persons are not equal therefore
the labour market is not perfect”. Test the validity of this
argument using truth table.
Ans. Refer Q. 4.11.
Q. 4. What is a tautology, contradiction and contingency ? Show
that (p q) ( p r) (q r) is a tautology, contradiction
or contingency.
4–20 C (CS/IT-Sem-3)
Propositional Logic & Predicate Logic
Ans. Refer Q. 4.12.
Q. 5. Show that the premises “It is not sunny this afternoon and
it is colder than yesterday,” “We will go swimming only if it
is sunny,” “If we do not go swimming, then we will take a
canoe trip.” and “If we take a canoe trip, then we will be
home by sunset” lead to the conclusion “We will be home by
sunset.”
Ans. Refer Q. 4.13.
Q. 6. Show that : (r ~ q, r S, S ~ q, p q) ~ p are
inconsistent.
Ans. Refer Q. 4.14.
Q. 7. Write the s ymbolic form and negate the following
statements :
a. Everyone who is healthy can do all kinds of work.
b. Some people are not admired by everyone.
c. Everyone should help his neighbours, or his neighbours
will not help him.
Ans. Refer Q. 4.17.
Q. 8. Express the following statements using quantifiers and
logical connectives.
a. Mathematics book that is published in India has a blue
cover.
b. All animals are mortal. All human being are animal.
Therefore, all human being are mortal.
c. There exists a mathematics book with a cover that is not
blue.
d. He eats crackers only if he drinks milk.
e. There are mathematics books that are published outside
India.
f. Not all books have bibliographies.
Ans. Refer Q. 4.18.
Q. 9. Translate the following sentences in quantified expressions
of predicate logic.
i. All students need financial aid.
ii. Some cows are not white.
iii. Suresh will get if division if and only if he gets first div.
iv. If water is hot, then Shyam will swim in pool.
v. All integers are either even or odd integer.
Ans. Refer Q. 4.20.
Discrete Structures & Theory of Logic
5
UNIT
5–1 C (CS/IT-Sem-3)
Trees, Graph and
Combinatorics
CONTENTS
Part-1
:
Trees : Definition, ...................................... 5–2C to 5–4C
Binary Tree
Part-2
:
Binary Tree Traversal .............................. 5–5C to 5–8C
Part-3
:
Binary Search Tree ................................. 5–8C to 5–12C
Part-4
:
Graphs : Definition ................................ 5–12C to 5–15C
and Terminology
Representation of Graph
Part-5
:
Multigraph Bipartite .............................. 5–15C to 5–24C
Graphs, Planar Graph
Isomorphism and
Homomorphism of Graphs
Part-6
:
Graph Coloring ....................................... 5–24C to 5–24C
Part-7
:
Recurrence Relation and ..................... 5–24C to 5–36C
Generating Function :
Recursive Definition of
Function, Recursive
Algorithms, Method of Solving
Recurrences, Combinatorics :
Introduction, Counting Techniques,
Pigeonhole Principle
5–2 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
PART-1
Trees : Definition, Binary Tree.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.1.
i.
ii.
iii.
iv.
v.
Explain the following terms :
Tree
Forest
Binary tree
Complete binary tree
Full binary tree
Answer
i.
Tree :
1.
A tree is a connected graph that contains no cycle or circuit. It is a
simple graph having no self loop or parallel edges.
2.
A tree contains a finite set of elements which are called vertices or
nodes. The vertex can have minimum degree 1 and maximum
degree n.
The number of vertices in a tree is called order of tree.
A tree with only one vertex is called trivial or degenerate or empty
tree.
3.
4.
Fig. 5.1.1.
ii.
Forest : A forest is a collection of disjoint tree. It is an undirected,
disconnected, acyclic graph.
Fig. 5.1.2.
iii. Binary tree : Binary tree is the tree in which the degree of every
node is less than or equal to 2. A tree consisting of no nodes is also a
binary tree.
5–3 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
iv. Complete binary tree : A binary tree is said to be complete binary
tree if all its levels, except the last, have maximum number of possible
nodes (i.e., 2), and if all the nodes at the last level appear as far left as
possible.
a
b
c
e
d
f
g
h
i
j
Fig. 5.1.3.
It is represented as Tn where n is number of nodes.
v.
Full binary tree : A binary tree is said to be extended or full binary tree
if each node has either 0 or 2 children.
a
c
b
e
d
g
f
Fig. 5.1.4.
Que 5.2.
Consider the tree given below :
a
c
b
j
d
e
g
f
h
i
Fig. 5.2.1.
i.
Which node is root ?
ii.
Which nodes are leaves ?
iii. Name parent node of each ?
iv. List children of each node.
k
l
5–4 C (CS/IT-Sem-3)
v.
Trees, Graph and Combinatorics
List the siblings.
vi. Find depth of each node.
vii. Find level of each node.
Answer
i.
The node a is the root node.
ii.
The node g, h, i, j and l are leaves.
iii.
iv.
v.
Nodes
Parent node
b, c
d
j, k
e, f
g, h
i
l
a
b
c
d
e
f
k
Nodes
Children
a
b
c
d
e
f
k
b, c
d
j, k
e, f
g, h
i
l
b and c
e and f
g and h
j and k
vi.
vii.
Nodes
Depth
a
b, c
d, j, k
e, f, l
g, h, i
1
2
3
4
5
Nodes
Level
a
b, c
d, j, k
e, f, l
g, h, i
0
1
2
3
4
5–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
PART-2
Binary Tree Traversal.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.3.
Define preorder, inorder, and postorder tree traversal.
Give an example of preorder, postorder, and inorder traversal of a
binary tree of your choice with at least 12 vertices.
OR
Explain in detail about the binary tree traversal with an example.
AKTU 2016-17, Marks 10
Answer
Tree traversal : A traversal of tree is a process in which each vertex is
visited exactly once in a certain manner. For a binary tree we have three
types of traversal :
1. Preorder traversal : Each vertex is visited in the following order :
a. Visit the root (N).
b. Visit the left child (or subtree) of root (L).
c. Visit the right child (or subtree) of root (R).
2. Postorder traversal :
a. Visit the left child (subtree) of root.
b. Visit the right child (subtree) of root.
c. Visit the root.
3. Inorder traversal :
a. Visit the left child (subtree) of root.
b. Visit the root.
c. Visit the right child (subtree) of root.
A binary tree with 12 vertices :
A
C
B
D
H
G
E
F
I
J
K
Fig. 5.3.1.
L
5–6 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Preorder (NLR) : A B D H I E J C F K L G
Inorder (LNR)
: HDIBJEAKFLCG
Postorder (LRN) : H I D J E B K L F G C A
Que 5.4.
Construct a tree whose inorder and preorder traversal
are as follows :
Inorder
:
Q
B
A
G
C
P
E
D
R
Preorder :
G
B
Q
A
P
C
D
E
R
Answer
Root is G.
G
Fig. 5.4.1.
Element to left of G in inorder traversal are Q B A and B comes first in
preorder traversal. Therefore tree will be
G
B
Q
A
Fig. 5.4.2.
Now elements on the right of G in inorder traversal are CPEDR and P comes
first. Therefore tree will be
G
B
Q
P
A
Fig. 5.4.3.
Now element to left of P in inorder traversal is C and to the right is EDR
(right subtree) out of DER, D comes first and E and R on its left and right
respectively.
5–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Therefore, the complete binary will look like
G
P
B
Q
C
A
D
E
R
Fig. 5.4.4.
Que 5.5.
Define a binary tree. A binary tree has 11 nodes. It’s
inorder and preorder traversals node sequences are :
Preorder : A B D H I E J L C F G
Inorder : H D I B J E K A F C G
AKTU 2018-19, Marks 07
Draw the tree.
Answer
Binary tree : Refer Q. 5.1(iii), Page 5–2C, Unit-5.
Numerical :
Step 1 : In preorder sequence, leftmost element is the root of the tree. By
searching A in ‘Inorder Sequence’ we can find out all the elements on the left
and right sides of ‘A’.
A
FCG
HDIBJEK
Step 2 : We recursively follow the above steps and we get
A
B
C
A
B
HDI JEK
Que 5.6.
D
C
F
G
H
E
I
J
F
G
K
Given the inorder and postorder traversal of a tree T :
Inorder : HFEABIGDC
Postorder : BEHFACDGI
Determine the tree T and it's Preorder.
AKTU 2017-18, Marks 07
5–8 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Answer
The root of tree is I.
I
Now elements on right of I are D, G, C and G comes last of all in postorder
traversal.
I
G
Now D and C are on right of G and D comes last of G and I postorder
traversal so
I
G
D
C
Now element left of I are H F E A B in inorder traversal and A comes last of
all in postorder traversal. Therefore tree will be
I
G
A
D
C
Now H F E are on left of A in inorder traversal and B comes last of all and
continuing in same manner. We will get final binary tree as
I
G
A
H
D
B
F
C
E
Preorder traversal of above binary tree is CAFHEBDGI
PART-3
Binary Search Tree.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.7.
What is a binary search tree ? Form a binary search
tree for the words vireo, warbler, egret, grosbeak, nuthatch, and
kingfisher. Explain each step.
5–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Answer
Binary search tree :
1.
A binary search tree is a binary tree T in which data is associated with
the vertices.
2.
The data are arranged so that, for each vertex v in T, each data item in
the left subtree of v is less than the data item in v and each data item
in the right subtree of v is greater than the data item in v.
3.
The binary tree T in Fig. 5.7.1, is a binary search tree since every
vertex in T exceeds every number in its left subtree and is less than
every number in its right subtree.
50
40
70
30
75
45
65
42
Fig. 5.7.1. A binary search tree.
Numerical :
1.
To start, create a vertex and place the first item in the list in this vertex
and assign this as the key of the root.
vireo
2.
To add warbler, compare it with vireo. Warbler is greater than vireo,
therefore it will be in the right of vireo. i.e.,
vireo
warbler
3.
Similarly by doing the above steps for rest of words we get the final
binary search tree that is given below :
vireo
egret
warbler
grosbeak
nuthatch
kingfisher
5–10 C (CS/IT-Sem-3)
Que 5.8.
i.
ii.
Trees, Graph and Combinatorics
Write algorithm for following :
Searching and inserting a node in BST
Deleting a node in BST
Answer
i.
Searching and inserting a node in BST :
Consider a binary search tree T and we have to insert or search an ITEM in
the tree.
Step I : Compare ITEM with the root of the tree
a.
If ITEM < root, proceed to the left child of N.
b.
If ITEM > root, proceed to the right child of N.
Step II : Repeat Step I and if
a.
We reach a node N such that ITEM = N then search is successful.
b.
We reach an empty subtree, which shows the search is successful.
Insert ITEM in place of empty subtree.
ii.
Deleting a node in BST :
Consider a binary search tree T and we have to delete an ITEM from the
tree.
Step I : Find the location of node which contain ITEM and also keep track of
its parent too.
Step II : Find the number of children of the node.
a.
If node has no children then simply delete the node.
b.
If the node has exactly one child then the node is deleted from tree by
replacing the node from its child.
c.
If the node has two children then
i.
Find its inorder successor.
ii. Delete the successor from tree using (a) or (b).
iii. Replace node by its successor in tree.
Que 5.9.
Draw a binary search tree by inserting following
integers 55, 20, 63, 10, 28, 60, 93, 5, 11, 40, 68, 25.
Answer
1.
Insert 55 :
2.
Insert 20 :
55
55
20
5–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
3.
Insert 63 :
55
20
4.
63
Insert 10 :
55
20
63
10
5.
Insert 28 :
55
63
20
28
10
6.
Insert 60 :
55
63
20
10
7.
28
60
Insert 93 :
55
63
20
10
8.
28
60
93
Insert 5 :
55
63
20
10
5
28
60
93
5–12 C (CS/IT-Sem-3)
9.
Trees, Graph and Combinatorics
Insert 11 :
55
63
20
10
5
28
60
93
11
10. Insert 40 :
55
63
20
10
5
28
11
60
93
40
11. Insert 68 :
55
63
20
10
5
28
11
60
40
93
68
12. Insert 25 :
55
63
20
10
5
28
11
25
60
40
93
68
PART-4
Graphs : Definition and Terminology, Representation of Graph.
5–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.10. What do you mean by graph ? Also, explain directed and
undirected graph.
Answer
A graph is a non-linear data structure consisting of nodes and edges. A graph
consists of two sets as follows :
1. Set V of nodes or point or vertices of graph G.
2. Set E of ordered or unordered pairs of distinct edges of G.
We denote such a graph by G(V, E) and set of vertices as V(G) and set of
edges as E(G).
For example :
V4
c
V3
d
e
b
V1
V2
a
Fig. 5.10.1.
Order : If G is finite then number of vertices in G denoted by |V (G)| is
called order of G.
Size : The number of edges denoted by |E(G)|in a finite graph G is called
size of G.
Directed graph : A graph G(V, E) is said to be directed graph or digraph if
each edge e E is associated with an ordered pair of vertices as shown
below :
c
a
b
Fig. 5.10.2.
Undirected graph : A graph G(V, E) is said to be undirected if each edge
e E is associated with an unordered pair of vertices as shown below :
a
b
c
d
Fig. 5.10.3.
5–14 C (CS/IT-Sem-3)
Que 5.11.
Trees, Graph and Combinatorics
Discuss representation of graph.
Answer
Graph can be represented in following two ways :
1. Matrix representation :
Matrices are commonly used to represent graphs for computer
processing. Advantages of representing the graph in matrix lies in the
fact that many results of matrix algebra can be readily applied to study
the structural properties of graph from an algebraic point of view.
a. Adjacency matrix :
i.
Representation of undirected graph :
The adjacency matrix of a graph G with n vertices and no
parallel edges is a n × n matrix A = [aij] whose elements are
given by
aij = 1, if there is an edge between ith and jth vertices
= 0, if there is no edge between them
ii. Representation of directed graph :
The adjacency matrix of a digraph D, with n vertices is the
matrix
A = [aij]n×n in which
aij = 1 if arc (vi, vj) is in D
= 0 otherwise
For example :
V1
V4
V2
V3
Fig. 5.11.1.
v1
b.
v1 0
A = v2 1
v3 1
v4 1
Incidence matrix :
i.
v2
v3
v4
1 1 1
0 1 0
1 0 1
0 1 0
Representation of undirected graph :
Consider an undirected graph G = (V, E) which has n vertices
and m edges all labelled. The incidence matrix I(G) = [bij], is
then n × m matrix, where
bij = 1 when edge ej is incident with vi
= 0 otherwise
5–15 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
ii.
Representation of directed graph :
The incidence matrix I(D) = [bij] of digraph D with n vertices
and m edges is the n × m matrix in which.
bij = 1 if arc j is directed away from a vertex vi
= – 1 if arc j is directed towards vertex vi
= 0 otherwise.
Find the incidence matrix to represent the graph shown in
Fig. 5.11.2.
v1
e1
e4
v4
e5
e3
v2
e2
v3
Fig. 5.11.2.
The incidence matrix of the digraph of Fig. 5.11.2 is
2.
0
0 1
1
1
1
1
0
0
0
I(D) =
0 1
1
0 1
0 1
1
0
0
Linked representation : In this representation, a list of vertices
adjacent to each vertex is maintained. This representation is also called
adjacency structure representation. In case of a directed graph, a care
has to be taken, according to the direction of an edge, while placing a
vertex in the adjacent structure representation of another vertex.
PART-5
Multigraph, Bipartite Graphs, Planar Graph,
Isomorphism and Homomorphism of Graphs.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.12.
a.
b.
Write short notes on :
Simple and multigraph
Complete graph and regular graph
5–16 C (CS/IT-Sem-3)
c.
d.
Trees, Graph and Combinatorics
Bipartite graph
Planar graph
Answer
a.
Simple and multigraph :
i.
Simple graph : A graph in which there is only one edge between
a pair of vertices is called a simple graph.
e
V1
V2
Fig. 5.12.1.
ii.
Multigraph : Any graph which contains some parallel edges is
called a multigraph.
V4
e3
e4
V1
V3
e2
e1
V2
Fig. 5.12.2.
b.
Complete graph and regular graph :
i.
Complete graph : A simple graph, in which there is exactly one
edge between each pair of distinct vertices is called a complete
graph. The complete graph of n vertices is denoted by Kn. The
graphs K1 to K5 are shown below in Fig. 5.12.3.
K1
K2
K3
K4
K5
Fig. 5.12.3.
ii.
n(n – 1) n
Kn has exactly
= C2 edges
2
Regular graph (n-regular graph) : If every vertex of a simple
graph has equal edges then it is called regular graph.
If the degree of each vertex is n then the graph is called n-regular
graph.
(a)
(a)
Fig. 5.12.4.
(c)
5–17 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
The graphs shown in Fig. 5.12.4 are 2-regular graphs.
The graph shown in Fig. 5.12.5 is 3-regular graph.
Fig. 5.12.5.
c.
Bipartite graph :
i.
Bipartite graph : A graph G = (V, E) is bipartite if the vertex set
V can be partitioned into two subsets (disjoint) V1 and V2 such that
every edge in E connects a vertex in V1 and a vertex V2 (so that no
edge in G connects either two vertices in V1 or two vertices in V2).
(V1, V2) is called a bipartition of G.
x1
x2
x3
x4
y1
y2
x1
x2
x3
y1
y2
y3
x4
which
redrawn
as :
y3
Fig. 5.12.6. Some bipartite graphs.
ii.
Complete bipartite graph : The complete bipartite graph on m
and n vertices, denoted Km, n is the graph, whose vertex set is
partitioned into sets V1 with m vertices and V2 with n vertices in
which there is an edge between each pair of vertices 1 and 2
where 1 is in V1 and 2 is in V2. The complete bipartite graphs
K2,3, K2,4, K 3,3, K3,5, and K2,6
K2,3
K3,5
K2,4
K3,3
K2,6
Fig. 5.12.7. Some complete bipartite graphs.
d.
Planar graph :
A graph G is said to be planar if there exists some geometric
representation of G which can be drawn on a plane such that no two of
its edges intersect except only at the common vertex.
i.
A graph is said a planar graph, if it cannot be drawn on a plane
without a crossover between its edges crossing.
ii.
The graphs shown in Fig. 5.12.8(a) and (b) are planar graphs.
5–18 C (CS/IT-Sem-3)
(a)
Que 5.13.
Trees, Graph and Combinatorics
(b )
Fig. 5.12.8. Some planar graph.
Define the following with one example :
i.
Bipartite graph
ii. Complete graph
iii. How many edges in K7 and K3, 6
AKTU 2017-18, Marks 10
iv. Planar graph
Answer
i.
Refer Q. 5.12(c), Page 5–15C, Unit-5.
ii.
Refer Q. 5.12(b), Page 5–15C, Unit-5.
iii. Number of edge in K7 : Since, Kn is complete graph with n vertices.
7(7 1) 7 6
= 21
2
2
Number of edge in K3, 6 :
Number of edge in K7 =
Since, Kn, m is a complete bipartite graph with n V1 and m V2
Number of edge in K3, 6 = 3 × 6 = 18
iv. Refer Q. 5.15(d), Page 5–18A, Unit-5.
Que 5.14.
Explain isomorphism and homomorphism of graph.
Answer
Isomorphism of graph : Two graphs are isomorphic to each other if :
i.
Both have same number of vertices and edges.
ii.
Degree sequence of both graphs are same (degree sequence is the
sequence of degrees of the vertices of a graph arranged in non-increasing
order).
Example :
Fig. 5.14.1.
Homomorphism of graph : Two graphs are said to be homomorphic if one
graph can be obtained from the other by the creation of edges in series (i.e.,
by insertion of vertices of degree two) or by the merger of edges in series.
5–19 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
(a)
(b )
Fig. 5.14.2.
(c )
Que 5.15. Prove that K3 and K4 are planar graphs. Prove that K5 is
non-planar.
Answer
The complete K3 graph has 3 edges and 3 vertices.
For a graph to be planar 3 – e 6
3 – e = 3 × 3 – 3 = 9 – 3 = 6 6
K3 is planar graph
Similarly complete K4 graph has 4 vertices and 6 edges.
3 – e = 3 × 4 – 6 = 12 – 6 = 6 6
K4 is planar graph
The complete K5 graph contains 5 vertices and 10 edges.
Now 3 – e = 3 × 5 – 10 = 15 – 10 = 5 6
Hence K5 is non planar since for a graph to be planar 3v – e 6.
Que 5.16.
What are Euler and Hamiltonian graph ?
OR
Explain the following terms with example :
i.
Homomorphism and Isomorphism graph
ii. Euler graph and Hamiltonian graph
iii. Planar and Complete bipartite graph
AKTU 2014-15, Marks 10
Answer
i.
ii.
Refer Q. 5.14, Page 5–18C, Unit-5.
Eulerian path : A path of graph G which includes each edge of G
exactly once is called Eulerian path.
Eulerian circuit : A circuit of graph G which include each edge of G exactly
once.
Eulerain graph : A graph containing an Eulerian circuit is called Eulerian
graph.
For example : Graphs given below are Eulerian graphs.
V1
V5
V2
V3
A
B
V4
D
C
Fig. 5.16.1.
5–20 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Hamiltonian graph : A Hamiltonian circuit in a graph G is a closed path
that visit every vertex in G exactly once except the end vertices. A graph G is
called Hamiltonian graph if it contains a Hamiltonian circuit.
For example : Consider graphs given below :
C
D
C
D
F
A
B
E
A
(a)
B
(b )
Fig. 5.16.2.
Graph given is Fig. 5.16.2(a) is a Hamiltonian graph since it contains a
Hamiltonian circuit A – B – C – D – A while graph in Fig 5.15.2(b) is not a
Hamiltonian graph.
Hamiltonian path : The path obtained by removing any one edge from a
Hamiltonian circuit is called Hamiltonian path. Hamiltonian path is subgraph
of Hamiltonian circuit. But converse is not true.
The length of Hamiltonian path in a connected graph of n vertices is
n – 1 if it exists.
iii. Refer Q. 5.12, Page 5–15C, Unit-5.
Que 5.17.
a.
b.
Prove that a connected graph G is Euler graph if and only if
every vertex of G is of even degree.
Which of the following simple graph have a Hamiltonian circuit
AKTU 2016-17, Marks 15
or, if not a Hamiltonian path ?
a
b
a
b
e
c
d
c
d
G1
G2
a
b
d
c
g
e
f
G3
Fig. 5.17.1.
Answer
a.
1.
First of all we shall prove that if a non-empty connected graph is
Eulerian then it has no vertices of odd degree.
2.
Let G be Eulerian.
3.
Then G has an Eulerian trail which begins and ends at u.
Discrete Structures & Theory of Logic
5–21 C (CS/IT-Sem-3)
4.
If we travel along the trail then each time we visit a vertex. We use
two edges, one in and one out.
5. This is also true for the start vertex because we also end there.
6. Since an Eulerian trail uses every edge once, the degree of each
vertex must be a multiple of two and hence there are no vertices of
odd degree.
7. Now we shall prove that if a non-empty connected graph has no
vertices of odd degree then it is Eulerian.
8. Let every vertex of G have even degree.
9. We will now use a proof by mathematical induction on |E(G)|, the
number of edges of G.
Basis of induction :
Let |E(G)| = 0, then G is the graph K1, and G is Eulerian.
Inductive step :
1. Let P(n) be the statement that all connected graphs on n edges of
even degree are Eulerian.
2. Assume P(n) is true for all n < |E(G)|.
3. Since each vertex has degree at least two, G contains a cycle C.
4. Delete the edges of the cycle C from G.
5. The resulting graph, G say, may not be connected.
6. However, each of its components will be connected, and will have
fewer than |E(G)| edges.
7. Also, all vertices of each component will be of even degree, because
the removal of the cycle either leaves the degree of a vertex
unchanged, or reduces it by two.
8. By the induction assumption, each component of G is therefore
Eulerian.
9. To show that G has an Eulerian trail, we start the trail at a vertex,
u say, of the cycle C and traverse the cycle until we meet a vertex,
c1 say, of one of the components of G .
10. We then traverse that component’s Eulerian trail, finally returning
to the cycle C at the same vertex, c1.
11. We then continue along the cycle C, traversing each component of
G as it meets the cycle.
12. Eventually, this process traverses all the edges of G and arrives
back at u, thus producing an Eulerian trail for G.
13. Thus, G is Eulerian by the principle of mathematical induction.
b.
G1 : The graph G1 shown in Fig. 5.17.1 contains Hamiltonian circuit, i.e.,
a – b – c – d – e – a and also a Hamiltonian path, i.e., abcde.
G2 : The graph G2 shown in Fig. 5.17.1 does not contain Hamiltonian
circuit since every cycle containing every vertex must contain the edge
e twice. But the graph does have a Hamiltonian path a – b – c – d.
5–22 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
G3 : The graph G3 shown in Fig. 5.17.1 neither have Hamiltonian circuit
nor have Hamiltonian path because any traversal does not cover all the
vertices.
Que 5.18.
Prove that a simple graph with n vertices and k
(n k) ( n k 1)
edges.
2
components can have at most
AKTU 2016-17, Marks 10
Answer
Let the number of vertices in each of the k-components of a graph G be n1,
n2, ...., nk, then we get
n1 + n2 + ... + nk = n where ni 1 (i = 1, 2, ..., k)
k
k
Now,
1) =
(n
i
k
n 1 = n – k
i
i 1
i 1
i 1
2
k
2
2
(ni 1) = n + k – 2nk
i 1
k
or
k
(n
i
i 1
1)2 2 (ni 1)(n j 1) = n2 + k2 – 2nk
i 1 j 1
i j
k
or
(n
i
1)2 + 2(non-negative terms) = n2 + k2 – 2nk
i 1
[ ni – 1 0, nj – 1 0]
k
1)2 n2 + k2 – 2nk
(n
or
i
i 1
k
or
k
k
2
i
n 1 2 n
i 1
i
i 1
n2 + k2 – 2nk
i 1
k
or
2
i
n
k 2n n2 + k2 – 2nk
i 1
k
or
2
i
n
n n2 + k2 – 2nk – k + n
i 1
= n(n – k + 1) – k (n – k + 1)
= (n – k)(n – k + 1)
...(5.18.1)
We know that the maximum number of edges in the ith component of
ni (ni 1)
2
Therefore, the maximum number of edges in G is :
G=
ni
C2 =
1
1
ni (ni 1) = 2
2
n n
2
i
i
=
1
2
n
2
i
n
Discrete Structures & Theory of Logic
Que 5.19.
5–23 C (CS/IT-Sem-3)
1
(n – k)(n – k + 1) by using eq. (5.18.1)
2
What are different ways to represent a graph ? Define
Euler circuit and Euler graph. Give necessary and sufficient
conditions for Euler circuits and paths. AKTU 2018-19, Marks 07
Answer
Representation of graph : Refer Q. 5.11, Page 5–14C, Unit-5.
Euler circuit and Euler graph : Refer Q. 5.16, Page 5–19C, Unit-5.
Necessary and sufficient condition for Euler circuits and paths :
1.
2.
A graph has an Euler circuit if and only if the degree of every vertex is
even.
A graph has an Euler path if and only if there are at most two vertices
with odd degree.
Que 5.20.
Define and explain any two the following :
1.
2.
BFS and DFS in trees
Euler graph
3.
Adjacency matrix of a graph
AKTU 2017-18, Marks 07
Answer
1.
Breadth First Search (BFS) : Breadth First Search (BFS) is an
algorithm for traversing or searching tree or graph data structures. It
starts at the tree root and explores the neighbour nodes first, before
moving to the next level neighbours.
Algorithmic steps :
Step 1 : Push the root node in the queue.
Step 2 : Loop until the queue is empty.
Step 3 : Remove the node from the queue.
Step 4 : If the removed node has unvisited child nodes, mark them as
visited and insert the unvisited children in the queue.
Depth First Search (DFS) :
Depth First Search (DFS) is an algorithm for traversing or searching
tree or graph data structures. One starts at the root (selecting some
arbitrary node as the root in the case of a graph) and explores as far as
possible along each branch before backtracking.
Algorithmic steps :
Step 1 : Push the root node in the stack.
Step 2 : Loop until stack is empty.
Step 3 : Pick the node of the stack.
Step 4 : If the node has unvisited child nodes, get the unvisited child
node, mark it as traversed and push it on stack.
5–24 C (CS/IT-Sem-3)
2.
3.
Trees, Graph and Combinatorics
Step 5 : If the node does not have any unvisited child nodes, pop the
node from the stack.
Refer Q. 5.16, Page 5–19C, Unit-5.
Refer Q. 5.11, Page 5–14C, Unit-5.
PART-6
Graph Coloring.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.21.
Write a short note on graph coloring.
Answer
1.
Suppose that G (V, E) is a graph with no multiple edges, a vertex colouring
of G is an assignment of colours.
2.
A graph G is m-colourable if there exists a colouring of G which uses m
colours.
3.
Colouring the vertices such a way such that no two adjacent vertices
have same colour is called proper colouring otherwise it is called
improper colouring.
PART-7
Recurrence Relation and Generating Function : Recursive
Definition of Function, Recursive Algorithms, Method of
Solving Recurrences, Combinatorics : Introduction,
Counting Techniques, Pigeonhole Principle.
Questions-Answers
Long Answer Type and Medium Answer Type Questions
Que 5.22.
Write a short note on recurrence relation.
Answer
For a sequence of numbers or numeric function (a0, a1, a2,... ar, ....) an
equation relating ar, for any r, to one or more of the ais, i < r is called
5–25 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
recurrence relation. Recurrence relations are also called difference equations
because they can be written in terms of the difference between the
consecutive terms of a sequence.
For example :
ar = 4ar – 1
r1
ar = ar – 1 – 2ar – 2
r2
are recurrence relations.
Order of recurrence relation : The difference between the highest and
lowest subscripts of ar or f(x) or yn is called the order of recurrence relation.
For example :
1. The equation 13ar + 12ar – 1 = 0 is of order 1 (i.e., r – (r – 1) = 1).
2. The equation 8 f(x) + 2 f(x + 1) + f(x + 2) = k(x) is of order 2
(i.e., (x + 2) – x = 2).
Degree of recurrence relation : The highest power of ar or f(x) or yn is
called degree of recurrence relation.
For example :
1.
The equation y3k + 2 + 2y2k + 1 – yk = 0 has the degree 3 as the highest
power of yk is 3.
2.
The equation a4r + 2a3r – 1 + 3a2r – 2 + 4ar – 3 = 0 has the degree 4, as the
highest power of ar is 4.
Que 5.23.
What do you mean generating function ? Solve the
recurrence relation :
an = 2 an–1 – an–2, n 2 given a0 = 3, a1 = – 2
using generating function.
Answer
Generating function : The generating function for the sequence a0, a1, ...
ak, ... of real numbers is infinite series given by
G(x) = a0 + a1x + a2x2 + ...... + akxk + ...... =
a x
k
k
k0
x is considered just a symbol called indeterminate and it is not variable, which
is replaced by numbers belonging to same domain.
The given recurrence relation is,
an = 2an–1 – an–2, n 2
Multiply by xn and take summation from n = 2 to , we get
a x
n
n 2
n
...(5.23.1)
n
n
= 2 an 1 x an 2 x
n2
...(5.23.2)
n 2
We know,
G(x) =
a
n
x n a0 a1 x a2 x 2 a3 x3 ...
...(5.23.3)
n 0
From eq. (5.23.2), we have
(a2x2 + a3 x3+ ...) = 2 (a1x2 + a2 x3 + ...) – (a0 x2 + a1x3 + ....)
(a2x2 + a3 x3+ ...) = 2x (a1x + a2 x2 + ...) – x2 (a0 + a1 x + ....)
5–26 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Using eq. (5.23.3), we get
G(x) – a0 – a1x = 2x (G(x) – a0) – x2 G(x)
G(x) – 3 + 2x = 2x (G(x) – 3) – x2 G(x)
G(x)[1 – 2x + x2] = 3 – 8x
3
8x
3 8x
3 8x
3 8x
=
(1 x)2 (1 x)2
x 2 2 x 1 ( x 1)2 (1 x)2
an= 3(n + 1) – 8n = 3 – 5n
G(x) =
Solve the recurrence relation yn + 2 – 5yn + 1 + 6yn = 5n
Que 5.24.
AKTU 2016-17, Marks 10
subject to the condition y0 = 0, y1 = 2.
Answer
Let G(t) =
a t
n
n
be generating function of sequence {an}.
n 0
Multiplying given equation by tn and summing from
n = 0 to , we have
a
n2
tn – 5
n 0
t2
Put
n 1
n 0
G(t) – a0 – a1 t
a
tn 6
a
n
tn
n 0
5
n
tn
n 0
1
G (t) – a0
– 5
6 G(t) 1 – 5t
t
a0 = 0 and a1 = 2
G(t) – 2t – 5t G(t) + 6t2 G(t) =
t2
1 – 5t
G(t) – 5t G(t) + 6t2 G(t) =
t2
+ 2t
1 – 5t
G(t) (1 – 5t + 6t2) =
t2
+ 2t
1 – 5t
(6t2 – 5t + 1) G(t) =
t2
+ 2t
1 – 5t
G(t) =
t2
2t
(1 – 5t)(3t – 1)(2t – 1) (3t – 1)(2t – 1)
=
t2
2t
(1 – 5t)(1 – 3t)(1 – 2t) (1 – 3t)(1 – 2t)
Let
A
B
C
t2
=
(1 – 5t) (1 – 3t) (1 – 2t)
(1 – 5t)(1 – 3t)(1 – 2t)
Discrete Structures & Theory of Logic
A = (1 – 5t)
5–27 C (CS/IT-Sem-3)
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/5
=
t2
(1 – 3t)(1 – 2t) t 1/5
=
1 / 25
1
=
(1 – 3 / 5)(1 – 2 / 5)
6
B = (1 – 3t)
=
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/3
1/9
t2
=
(1 – 5t)(1 – 2t) t 1/3 3 – 5 3 – 2
3 3
= –
1
2
C = (1 – 2t)
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/2
t2
1/ 4
=
(2 – 5) (2 – 3)
(1 – 5t)(1 – 3t) t 1/2
2
2
1
=
3
=
Again,
2t
D
E
=
(1 – 3t)(1 – 2t)
(1 – 3t) (1 – 2t)
2t
D = (1 – 3t)
(1 – 3t)(1 – 2t) t 1/3
2t
2/3
2
=
(3 – 2)
(1 – 2t) t 1/3
3
2t
E = (1 – 2t)
(1 – 3t)(1 – 2t) t 1/ 2
=
2t
2/2
–2
=
2–3
(1 – 3t) t 1/ 2
2
1/ 6
1/ 2
1/3
2
2
G(t) =
–
–
(1 – 5t) (1 – 3t) (1 – 2t) (1 – 3t) (1 – 2t)
=
5–28 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
=
tn =
a
n
n 0
1
6
(5t)
n
n 0
3
2
(3t)
n
n 0
–
5
(2t)n
3 n 0
1
3
5
an = (5)n (3) n – (2)n
6
2
3
Que 5.25.
1/6
3/2
5/3
–
1 – 5t (1 – 3t) 1 – 2t
Solve the recurrence relation by the method of
generating function :
ar – 7 ar–1 + 10ar–2 = 0, r 2. Given a0 = 3 and a1 = 3.
AKTU 2014-15, Marks 10
OR
Solve the recurrence relation using generating function :
an – 7an – 1 + 10an – 2 = 0 with a0 = 3, a1 = 3.
AKTU 2015-16, Marks 10
Answer
ar – 7ar–1 + 10 ar–2 = 0, r 2
Multiply by xr and take sum from 2 to .
r2
r2
r2
ar xr 7 ar –1 xr 10 ar 2 xr 0
(a2 x2 + a3 x3 + a4 x4 + ....) – 7 (a1 x2 + a2 x3 + ....)
+ 10 (a0 x2 + a1 x3 + ....) = 0
We know that
G(x) =
ar xr a0 a1 x ....
r0
G(x) – a0 – a1 x – 7 x (G(x) – a0) + 10 x2 G(x) = 0
G(x) [1 – 7 x + 10 x2] = a0 + a1 x – 7 a0 x
= 3 + 3x – 21 x = 3 – 18 x
G(x) =
=
3 18 x
3 18 x
10 x 2 7 x 1 10 x 2 5 x 2 x 1
3 18 x
3 18 x
5 x (2 x 1) 1 (2 x 1) (5 x – 1)(2 x 1)
Now,
3 18 x
A
B
=
(5 x 1) (2 x 1)
5 x 1 2 x 1
5–29 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
3 – 18 x = A (2 x – 1) + B (5 x – 1)
put
5
3
3 – 9 = B 1 – 6 =
B B = – 4
2
2
1
x=
5
3
3
2
18
3–
= A 1 A 1 A = 1
5
5
5
5
put
1
4
4
1
=
5 x 1 2x 1
1 2x 1 5x
ar = 4.2r – 5r
G(x) =
Que 5.26.
1
2
x=
Solve the recurrence relation ar+2 – 5ar+1 + 6 ar = (r + 1)2.
AKTU 2014-15, Marks 10
Answer
ar+2 – 5 ar+1 + 6 ar = (r + 1)2 = r2 + 2r + 1
...(5.26.1)
Now the characteristic equation is :
x2 – 5 x + 6 = 0
(x – 3) (x – 2) = 0 x = 3, 2
The homogeneous solution is :
ar(h) = C12r + C2 3r
Let the particular solution be :
ar(p) = A0 + A1r + A2r2
From eq. (5.26.1)
A0 + A1 (r + 2) + A2 (r + 2)2 – 5 {A0 + A1 (r + 1)} + A2 (r + 1)2}
+ 6A0 + 6A1r + 6A2 r2
=
r2
+2r+1
(A0 + 2A1 + 4A2 –5A0 – 5A1 – 5A2 + 6A0) + r (A1 + 4A2 – 5A1 – 10A2 + 6A1)
+ r2 ( A2 – 5A2 + 6A2) = r2 + 2r + 1
Comparing both sides, we get,
2A0 – 3A1 – A2 = 1
...(5.26.2)
2A1 – 6A2 = 2
2A2 = 1
From eq. (5.26.3), 2A1 – 3 = 2
A1 =
5
2
...(5.26.3)
A2 = 1/2
5–30 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
From eq. (5.26.2)
2A0 –
15 1
=1
2 2
2A0 – 8 = 1
ar(p) =
The final solution is,
Que 5.27.
A0 =
9
2
9 5
r2
r
2 2
2
ar = ar(h) + ar(p) = C1 2r + C2 3r +
9
5
r2
+ r
2
2
2
Solve ar – 6ar – 1 + 8ar – 2 = r.4r, given a0 = 8, and a1 = 1.
AKTU 2017-18, Marks 07
Answer
ar – 6ar – 1 + 8ar – 2 = r4r
The characteristic equation is, x2 – 6x + 8 = 0, x2 – 2x – 4x + 8 = 0
(x – 2) (x – 4) = 0, x = 2, 4
The solution of the associated non-homogeneous recurrence relation is,
ar(h) = B1(2)r + B2(4)r
...(5.27.1)
Let particular solution of given equation is, ar(p) = r2(A0 + A1r)4r
Substituting in the given equation, we get
r2(A0 + A1r)4r – 6(r – 1)2 (A0 + A1(r – 1))4r – 1
+ 8(r – 2)2 (A0 + A1(r – 2)4r – 2 = r4r
6
r 2A 0 + A 1r 3 –
[(A0r2 – 2A0r + A0) + (A1r3 – A1 – 3A1r2 + 3A1r)2]
4
8
+ 2 [(A0r2 – 4rA0 + 4A0) + (A1r3 – 8A1 – 6A1r\2 + 12A1r)] = r
4
3
3
3
3
rA0 + A1r3 –
A r2 + 3A0r –
A –
A r3 +
A
2 0
2 0 2 1
2 1
9
9
1
+
A r2 –
A r+
A r2 – 2A0r + 2A0
2 1
2 1
2 0
1
A r3 – 4A1 – 3A1r2 – 6A1r = r
2 1
1
5
3
3
2A0r – A0r2 –
A –
A +
A r2 +
A r=r
2 0 2 1 2 1
2 1
Comparing both sides, we get
3
2A0 +
A =1
...(5.27.2)
2 1
A0 + 5A1 = 0
...(5.27.3)
2
10
Solving equation (5.27.2) and (5.27.3), we get A1 =
A0 =
17
17
To find the value of B1 and B2 put r = 0 and r = 1 in equation (5.27.1)
Discrete Structures & Theory of Logic
r = 0 a0 = B1 + B2
5–31 C (CS/IT-Sem-3)
B1 + B2 = 8
r = 1 a1 = 2B1 + 4B2
...(5.27.4)
...(5.27.5)
31
15
Solving equations (5.27.4) and (5.27.5), we get B1 =
B2 =
2
2
Complete solution is, ar = ar(h) + ar(p)
10 2 r
31 r 15 r
ar =
2
4 r 2
r 4
2
2
17 17
Que 5.28.
2B1 + 4B2 = 1
Solve the recurrence relation : ar + 4ar – 2 + 4ar – 2 = r2.
AKTU 2017-18, Marks 07
Answer
ar + 4ar – 1 + 4ar – 2 = r2
The characteristic equation is,
x2 + 4x + 4 = 0
(x + 2)2 = 0
x = –2, – 2
The homogeneous solution is, a(h) = (A0 + A1r) (– 2)r
The particular solution be, a(p) = (A0 + A1r) r2
Put ar, ar – 1 and ar – 2 from a(p) in the given equation, we get
r2A0 + A1r3 + 4A0(r – 1)2 + 4A1(r – 1)3 + 4A0(r – 2)2 + 4A1(r – 2)3 = r2
A0(r2 + 4r2 – 8r + 4 + 4r2 – 16r + 16) +
A1(r3 + 4r3 – 4 – 12r2 + 12r + 4r3 – 32 – 24r2 + 48r) = r2
A0(9r2 – 24r + 20) + A1(9r3 – 48r2 + 60r – 36) = r2
Comparing the coefficient of same power of r, we get
9A0 – 48A1 = 1
20A0 – 36A1 = 0
...(5.28.1)
...(5.28.2)
3
5
Solving equation (5.28.1) and (5.28.2) A0 =
A1
53
159
The complete solution is,
3 5 2
ar = ar(p) + ar(h) = (A0 + A1r) (– 2)r +
r r
53 159
Que 5.29.
Suppose that a valid codeword is an n-digit number in
decimal notation containing an even number of 0’s. Let an denote
the number of valid codewords of length n satisfying the recurrence
relation an = 8an – 1 + 10n – 1 and the initial condition a1 = 9. Use
generating functions to find an explicit formula for an.
AKTU 2018-19, Marks 07
5–32 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Answer
Let G(x) =
an xn be the generating function of the sequence a0, a1, a2 ....
n 0
we sum both sides of the last equations starting with n = 1. To find that
G(x) – 1 =
an x n
(8a
n –1 x
n 1
= 8
n –1 x
n
n 1
10
n –1 n
x
n 1
an–1 x n–1 x
n 1
10
n –1 n –1
x
n 1
= 8x
10 n–1 x n )
a
= 8x
n
n 1
a x
n
n
x
n 0
10
n
xn
n 0
= 8xG(x) + x/(1 – 10x)
Therefore, we have
G(x) – 1 = 8xG(x) + x/(1 – 10x)
Expanding the right hand side of the equation into partial fractions gives
G(x) =
1 1
1
2 1 8 x 1 10 x
G(x) =
1
n n
8 x
2 n 0
This is equivalent to
=
n 0
an =
Que 5.30.
10
n 0
x
n n
1 n
(8 10n ) x n
2
1 n
(8 10n )
2
Define permutation and combination. Also, write
difference between them.
Answer
1.
Permutation refers to different ways of arranging a set of object in a
sequential order.
2.
The number of permutations of n different things taken r ( n) at a
time is denoted by p(n, r) or nPr.
3.
Combination refers to several ways of choosing items from a large set
of object.
5–33 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
4.
The number of combinations of n different thing taken r ( n) at a time
is denoted by C(n, r) or nCr.
The selection of two letters from three letters a, b, c are ab bc ca and thus,
the number of combinations of 3 letters taken 2 at a time is C(3, 2) = 3
Difference between a permutation and combination :
S. No.
Permutation
Combination
1.
Both selection and arrangement
are made.
Only selection is made.
2.
Ordering of the selected object
is essential.
Ordering of the selected object is
not essential.
3.
Multiple permutations can be
derive from combination.
Single combination is derive from
single permutation.
4.
nP
r
Que 5.31.
=
n!
(n r)!
nC
r
=
n!
r !(n r )!
Suppose that a cookie shop has four different kinds of
cookies. How many different ways can six cookies be chosen ?
AKTU 2016-17, Marks 15
Answer
As the order in which each cookie is chosen does not matter and each kind of
cookies can be chosen as many as 6 times, the number of ways these cookies
can be chosen is the number of 6-combination with repetition allowed from
a set with 4 distinct elements.
The number of ways to choose six cookies in the bakery shop is the number
of 6 combinations of a set with four elements.
C(4 + 6 – 1, 6) = C(9, 6)
Since
C(9, 6) = C(9, 3) = (9 · 8 · 7) / (1 · 2 · 3) = 84
Therefore, there are 84 different ways to choose the six cookies.
Que 5.32.
i.
Write short notes on the following :
Recursive algorithms
ii. Pigeonhole principle
Answer
i.
Recursive algorithm :
A function is called recursively defined if the function refers to itself,
and satisfies the following step :
a.
There must be certain arguments for which the function does not
refer to itself, these arguments are called initial values (Base values).
5–34 C (CS/IT-Sem-3)
b.
Trees, Graph and Combinatorics
These base values are used to define the other values of the function
by using the function recursively.
Example :
ar = 3 ar–1, r 1, a0 = 1
then
a1 = 3a0 = 3
a2 = 3a1 = 3 (3a0) = 32 a0 = 32
and so on.
ar = 3r, r 0
ii.
1.
2.
3.
4.
5.
6.
7.
Pigeonhole principle :
The pigeonhole principle is sometime useful in counting methods.
If n pigeons are assigned to m pigeonholes then at least one pigeonhole
contains two or more pigeons (m < n).
Proof :
Let m pigeonholes be numbered with the numbers 1 through m.
Beginning with the pigeon 1, each pigeon is assigned in order to the
pigeonholes with the same number.
Since m < n i.e., the number of pigeonhole is less than the number of
pigeons, n-m pigeons are left without having assigned a pigeonhole.
Thus, at least one pigeonhole will be assigned to a more than one pigeon.
We note that the pigeonhole principle tells us nothing about how to
locate the pigeonhole that contains two or more pigeons.
It only asserts the existence of a pigeon hole containing two or more
pigeons.
To apply the principle one has to decide which objects will play the role
of pigeon and which objects will play the role of pigeonholes.
Que 5.33. Find the number of integers between 1 and 250 that are
divisible by any of the integers 2, 3, 5, and 7.
Answer
1, 2, 3, -------------- 250
Number of integers between 1 and 250 that are divisible by 2 :
Quotient of last number 2 – quotient of first number 2
(250 2 ) – (1 2) = 125 – 0 = 125
Number of integers between 1 and 250 that are divisible by 3
(250 3) – (1 3) = 83 – 0 = 83
Number of integers between 1 and 250 that are divisible by 5
(250 5) – (1 5) = 50 – 0 = 50
Number of integers between 1 and 250 that are divisible by 7
(250 7) – (1 7) = 35 – 0 = 5
Total number of integers divisible by only 2, 3, 5, 7, individually are
125 + 83 + 50 + 35 = 293
Number of integers divisible by (2 and 3) = 41
Discrete Structures & Theory of Logic
5–35 C (CS/IT-Sem-3)
Number of integers divisible by (2 and 5) = 25
Number of integers divisible by (2 and 7) = 17
Number of integers divisible by (3 and 5) = 16
Number of integers divisible by (3 and 7) = 11
Number of integers divisible by (5 and 7) = 7
Number of integers divisible by (2, 3 and 5) = 8
Number of integers divisible by (2, 3, and 7) = 5
Number of integers divisible by (2, 5, and 7) = 3
Number of integers divisible by (3, 5 and 7) = 2
––––
135
––––
Number of integers between 1 and 250 that are divisible by any of the
integer (2, 3, 5 and 7) = 293 – 135 = 158.
Que 5.34.
How many different rooms are needed to assign 500
classes, if there are 45 different time periods during in the university
time table that are available ?
Answer
Using pigeonhole principle :
n 1
500 1
n = 500, m = 45 =
1 45 1
m
At least 12 rooms are needed.
Here
Que 5.35.
A total of 1232 student have taken a course in Spanish,
879 have taken a course in French, and 114 have taken a course in
Russian. Further 103 have taken courses in both Spanish and
French, 23 have taken courses in both Spanish and Russian, and 14
have taken courses in both French and Russian. If 2092 students
have taken least one of Spanish, French and Russian, how many
students have taken a course in all three languages ?
AKTU 2018-19, Marks 07
Answer
Let S be the set of students who have taken a course in Spanish, F be the set
of students who have taken a course in French, and R be the set of students
who have taken a course in Russian. Then, we have
|S| = 1232, |F| = 879, |R| = 114, |S F| = 103, |S R| = 23,
|S R| = 14, and |SF R| = 23.
Using the equation
|SF R| = |S|+ |F|+ |R| – |S F| – |S R| – |S R| + |S
F R|,
5–36 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
2092 = 1232 + 879 + 114 – 103 – 23 – 14 +|SF R|,
|SF R| = 7.
|S F R| = ?
|S F| = 103
|S| = 1232
|F| = 879
|S R| = 23
|F R| = 14
|R| = 114
|S F R| = 2092
Fig. 5.35.1.
VERY IMPORTANT QUESTIONS
Following questions are very important. These questions
may be asked in your SESSIONALS as well as
UNIVERSITY EXAMINATION.
Q. 1. Explain in detail about the binary tree traversal with an
example.
Ans. Refer Q. 5.3.
Q. 2. Define a binary tree. A binary tree has 11 nodes. It’s inorder
and preorder traversals node sequences are :
Preorder : A B D H I E J L C F G
Inorder : H D I B J E K A F C G
Draw the tree.
Ans. Refer Q. 5.5.
Q. 3. Given the inorder and postorder traversal of a tree T :
Inorder : HFEABIGDC Postorder : BEHFACDGI
Determine the tree T and it's Preorder.
Ans. Refer Q. 5.6.
5–37 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Q. 4.
i.
ii.
iii.
iv.
Define the following with one example :
Bipartite graph
Complete graph
How many edges in K7 and K3, 6
Planar graph
Ans. Refer Q. 5.13.
Q. 5.
i.
ii.
iii.
Explain the following terms with example :
Homomorphism and Isomorphism graph
Euler graph and Hamiltonian graph
Planar and Complete bipartite graph
Ans. Refer Q. 5.16.
Q. 6.
a. Prove that a connected graph G is Euler graph if and only if
every vertex of G is of even degree.
b. Which of the following simple graph have a Hamiltonian
circuit or, if not a Hamiltonian path ?
a
b
a
b
e
c
d
c
d
G1
G2
a
b
d
c
g
e
f
G3
Fig. 1.
Ans. Refer Q. 5.17.
Q. 7. Prove that a simple graph with n vertices and k components
can have at most
(n k) ( n k 1)
edges.
2
Ans. Refer Q. 5.18.
Q. 8. What are different ways to represent a graph ? Define Euler
circuit and Euler graph. Give necessary and sufficient
conditions for Euler circuits and paths.
Ans. Refer Q. 5.19.
Q. 9.
1.
2.
3.
Define and explain any two the following :
BFS and DFS in trees
Euler graph
Adjacency matrix of a graph
Ans. Refer Q. 5.20.
5–38 C (CS/IT-Sem-3)
Trees, Graph and Combinatorics
Q. 10. Solve the recurrence relation yn + 2 – 5yn + 1 + 6yn = 5n subject
to the condition y0 = 0, y1 = 2.
Ans. Refer Q. 5.24.
Q. 11. Solve the recurrence relation using generating function :
an – 7an – 1 + 10an – 2 = 0 with a0 = 3, a1 = 3.
Ans. Refer Q. 5.25.
Q. 12. Solve the recurrence relation ar+2 – 5ar+1 + 6 ar = (r + 1)2.
Ans. Refer Q. 5.26.
Q. 13. Solve ar – 6ar – 1 + 8ar – 2 = r.4r, given a0 = 8, and a1 = 1.
Ans. Refer Q. 5.27.
Q. 14. Solve the recurrence relation : ar + 4ar – 2 + 4ar – 2 = r2.
Ans. Refer Q. 5.28.
Q. 15. Suppose that a valid codeword is an n-digit number in
decimal notation containing an even number of 0’s. Let an
denote the number of valid codewords of length n satisfying
the recurrence relation an = 8an – 1 + 10n – 1 and the initial
condition a1 = 9. Use generating functions to find an explicit
formula for an.
Ans. Refer Q. 5.29.
Q. 16. Suppose that a cookie shop has four different kinds of
cookies. How many different ways can six cookies be
chosen ?
Ans. Refer Q. 5.31.
Q. 17. A total of 1232 student have taken a course in Spanish, 879
have taken a course in French, and 114 have taken a course
in Russian. Further 103 have taken courses in both Spanish
and French, 23 have taken courses in both Spanish and
Russian, and 14 have taken courses in both French and
Russian. If 2092 students have taken least one of Spanish,
French and Russian, how many students have taken a
course in all three languages ?
Ans. Refer Q. 5.35.
Discrete Structures & Theory of Logic
1
SQ–1 C (CS/IT-Sem-3)
Set Theory, Functions
and Natural Numbers
(2 Marks Questions)
1.1. What do you understand by partition of a set ?
Ans. A partition of a set A is a collection of non-empty subsets A1,
A2, ......, An called blocks, such that each element of A is in exactly
one of the blocks. i.e.,
i. A is the union of all subsets A1 A2 ..... An = A.
ii. The subsets are pairwise disjoint, Ai Aj = for i j.
1.2. Define transitive closure with suitable example.
Ans. The relation obtained by adding the least number of ordered pairs
to ensure transitivity is called the transitive closure of the relation.
The transitive closure of R is denoted by R+.
1.3. Let R be a relation on the set of natural numbers N, as
R = {(x, y) : x, y N, 3x + y = 19}. Find the domain and range of
R. Verify whether R is reflexive. AKTU 2016-17, Marks 02
Ans. By definition of relation,
R = {(1, 16), (2, 13), (3, 10), (4, 7), (5, 4), (6, 1)}
Domain = {1, 2, 3, 4, 5, 6}
Range = {16, 13, 10, 7, 4, 1}
R is not reflexive since (1, 1) R.
1.4. Show that the relation R on the set Z of integers given by
R = {(a, b) : 3 divides a – b}, is an equivalence relation.
AKTU 2016-17, Marks 02
Ans. Reflexive : a – a = 0 is divisible by 3
(a, a) R a Z
R is reflexive.
Symmetric : Let (a, b) R a – b is divisible by 3
– (a – b) is divisible by 3
b – a is divisible by 3
(b, a) R
R is symmetric.
Transitive : Let (a, b) R and (b, c) R
a – b is divisible by 3 and b – c is divisible by 3
SQ–2 C (CS/IT-Sem-3)
2 Marks Questions
Then a – b + b – c is divisible by 3
a – c is divisible by 3
(a, c) R
R is transitive.
Hence, R is equivalence relation.
1.5. Mention some properties of cartesian product.
Ans. For the four sets A, B, C and D
i.
ii.
iii.
iv.
(A B) × (C D) = (A × C) (B × D)
(A – B) × C = (A × C) – (B × C)
(A B) × C = (A × C) (B × C)
A × (B C) = (A × B) (A × C)
1.6. Discuss countable and uncountable sets.
Ans. A set which is either empty, finite or countably infinite is called
countable otherwise uncountable. Thus, A is countable if there
exists a one-to-one function F from N onto A.
1.7. Write a short note on composition of relation.
Ans. The composite of, say R and S denoted by RoS, is the relation
consisting of ordered pairs (a, c) when a A and c C, and for
which there exists an element b B such that (a, b) R and
(b, c) S.
1.8. How many symmetric and reflexive binary relations are
possible on a set S with cardinality n ?
Ans. There are 2n(n + 1)/2 symmetric binary relations and 2n(n – 1) reflexive
binary relations are possible on a set S with cardinality.
1.9. Let A = {a, {a}}. Determine whether the following statements
are true or false :
i. {a, {a}} P(A)
ii. {a, {a}} P(A)
iii. {{{a}}} P(A)
iv. {{{a}}} P(A)
Ans.
i.
ii.
iii.
iv.
False : {a, {a}} is not an element of A.
True : {a,{a}} is a subset of A.
False : {{{a}}} is not an element of A.
False : {{{a}}} is a subset of A.
1.10. Find out the cardinality of the following sets :
A = {x : x is weeks in a leap year}
B = {x : x is a positive divisor of 24 and not equal to zero}
C = {{{}}}
D = {{, {}}}
Discrete Structures & Theory of Logic
SQ–3 C (CS/IT-Sem-3)
Ans.
i. We know that there are 52 weeks in a leap year.
The cardinality of A is 52.
ii. The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
|B|= 8
iii. The set C has only one element.
|C| = 1
iv. The set D has only one element.
|D| = 1
1.11. If the function f : R R defined by f(x) = x2, find f–1 (4) and
f–1(– 4).
Ans.
f–1(4) = {x R : f (x) = 4}
= {x R : x2 = 4}
= {x R : x = ± 2} = {–2, 2}
f–1(– 4) = {x R : f (x) = – 4}
= {x R : x2 = – 4}
= {x R : x = 2 1 } = since 2 1
are imaginary numbers
1.12. Show that if set A has 3 elements, then we can have 26
symmetric relations on A.
AKTU 2015-16, Marks 02
Ans. Number of elements in set = 3
Number of symmetric relations if number of elements is
n = 2n(n + 1)/2
Here,
n=3
Number of symmetric relations
= 23(3 + 1)/2
= 23(4)/2
= 26
Hence proved.
1.13. If f : A B is one-to-one onto mapping, then prove that
f –1 : B A will be one-to-one onto mapping.
AKTU 2015-16, Marks 02
Ans. Proof : Here f : A B is one-to-one and onto.
a1, a2 A and b1, b2 B so that
b1 = f(a1), b2 = f (a2) and a1 = f–1(b1), a2 = f–1 (b2)
As f is one-to-one
f(a1) = f(a2) a1 = a2
b1 = b2 f–1(b1) = f–1(b2)
i.e.,
f–1(b1) = f–1(b2) b1 = b2
f–1 is one-to-one function.
As f is onto.
SQ–4 C (CS/IT-Sem-3)
2 Marks Questions
Every element of B is associated with a unique element of A i.e.,
for any a A is pre-image of some b B where b = f(a) a = f–1(b)
i.e., for b B, there exists f–1 image a A.
Hence, f–1 is onto.
1.14. Define multiset and power set. Determine the power set
A = {1, 2}.
AKTU 2015-16, Marks 02
Ans. Multiset : Multisets are sets where an element can occur as a
member more than once.
For example : A = { p, p, p, q, q, q, r, r, r, r}
B = { p, p, q, q, q, r}
are multisets.
Power set : A power set is a set of all subsets of the set.
The power set of A = {1, 2} is {{}, {1}, {2}}.
1.15. Define cardinality.
Ans. Cardinality of a set is defined as the total number of elements in a
finite set.
1.16. Define ordered pair.
Ans. An ordered pair is a pair of objects whose components occur in a
special order. It is written by listing the two components in the
specified order, separating them by a comma and enclosing the pair
in parenthesis. In the ordered pair (a, b), a is called the first
component and b, the second component.
1.17. Find the power set of each of these sets, where a and b are
distinct elements.
i. {a}
ii. {a, b}
iii. {, {}}
iv. {a, {a}}
AKTU 2018-19, Marks 02
Ans.
i.
ii.
iii.
iv.
Power set of {a} = {{}, {a}}
Power set of {a, b} = {{}, {a}, {b}, {a, b}}
Power set of {, {}} = {}
Power set of {a, {a}} = {{}, {a}, {{a}}, {a, {a}}}
1.18. Define injective, surjective and bijective function.
AKTU 2018-19, Marks 02
Ans.
1. One-to-one function (Injective function or injection) : Let
f : X Y then f is called one-to-one function if for distinct elements
of X there are distinct image in Y i.e., f is one-to-one iff
SQ–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
f(x1) = f(x2) implies x1 = x2 x1, x2, X
A
B
f
Fig. 1.18.1. One-to-one.
2. Onto function (Surjection or surjective function) : Let
f : X Y then f is called onto function iff for every element y Y
there is an element x X with f(x) = y or f is onto if Range (f) = Y.
X
Y
f
Fig. 1.18.2. Onto.
3. One-to-one onto function (Bijective function or bijection) :
A function which is both one-to-one and onto is called one-to-one
onto function or bijective function.
X
1
2
3
Y
f
a
b
c
Fig. 1.18.3. One-to-one onto.
1.19. Let A = (2, 4, 5, 7, 8) = B, aRb if and only if a + b <= 12. Find
AKTU 2017-18, Marks 02
relation matrix.
Ans. R = {(2, 4), (2, 5), (2, 7), (2, 8) (4, 2), (4, 5), (4, 7), (4, 8) (5, 2), (5, 4),
(5, 7), (7, 2), (7, 4), (7, 5), (8, 2), (8, 4), (2, 2), (4, 4), (5, 5)}
2
2 1
4 1
mij = 5 1
7 1
8 1
4
1
1
1
5
1
1
1
7
1
1
1
8
1
1
0
1 1 0 0
1 0 0 0
SQ–6 C (CS/IT-Sem-3)
2 Marks Questions
1.20. Define union and intersection of multiset and find for
A = [1, 1, 4, 2, 2, 3], B = [1, 2, 2, 6, 3, 3]
AKTU 2017-18, Marks 02
Ans. Union : Let A and B be two multisets. Then, A B, is the multiset
where the multiplicity of an element in the maximum of its
multiplicities in A and B.
Intersection : The intersection of A and B, A B, is the multiset
where the multiplicity of an element is the minimum of its
multiplicities in A and B.
Numerical :
A = {1, 1, 4, 2, 2, 3}, B = {1, 2, 2, 6, 3, 3}
Union :
A B = {1, 2, 3, 4, 6}
Intersection :
A B = {1, 2, 2, 3}
SQ–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
2
Algebraic Structures
(2 Marks Questions)
2.1. List the properties of cosets.
Ans. Let H be a subgroup of G and ‘a’ and ‘b’ belong to G. Then,
i.
ii.
iii.
iv.
a aH
aH = H iff a H
aH = bH
aH = bH iff a–1 b H
2.2. List types of permutation group.
Ans. Types of permutation group are :
i.
ii.
iii.
iv.
Identity permutation
Inverse permutation
Cyclic permutation
Even and odd permutation
2.3. If a and b are any two elements of group G then prove
(a * b)– 1 = (b– 1 * a– 1).
Ans. Consider (a * b) *
(b – 1
AKTU 2015-16, Marks 02
a – 1)
*
= a * (b * b – 1) * a – 1
= a * e * a– 1
= a * a–1= e
–
1
–
1
Also (b * a ) * (a * b) = b – 1 * (a – 1 * a) * b
= b–1 * e * b
= b–1 * b = e
Therefore (a * b) – 1 = b – 1 * a – 1 for any a, b G
2.4. What do you mean by integral domain ?
Ans. A ring (R, +, ·) is called an integral domain if,
i. It is commutative.
ii. It has multiplicative identity element.
iii. It is without zero divisors.
2.5. Define ring and give an example of a ring with zero divisors.
AKTU 2016-17, Marks 02
OR
SQ–8 C (CS/IT-Sem-3)
Define ring and field.
2 Marks Questions
AKTU 2018-19, Marks 02
OR
Define rings and write its properties.
AKTU 2017-18, Marks 02
Ans. Ring : A non-empty set R is a ring if it is equipped with two binary
i.
ii.
iii.
iv.
v.
vi.
i.
ii.
iii.
operations called addition and multiplication and denoted by '+'
and '.' respectively i.e., for all a, b R we have a + b R and a.b R
and it satisfies the following properties :
Addition is associative i.e.,
(a + b) + c = a + (b + c) a, b, c R
Addition is commutative i.e.,
a + b = b + a a, b R
There exists an element 0 R such that
0 + a = a = a + 0, a R
To each element a in R there exists an element – a in R such that
a + (– a) = 0
Multiplication is associative i.e.,
a.(b.c) = (a.b).c, a b, c R
Multiplication is distributive with respect to addition i.e., for all
a, b, c R,
Example of ring with zero divisors : R = {a set of 2 × 2
matrices}.
Field : A ring R with at least two elements is called a field if it has
following properties :
R is commutative
R has unity
R is such that each non-zero element possesses multiplicative
inverse.
For example : The rings of real numbers and complex numbers
are also fields.
2.6. Prove that if a2 = a, then a = e, a being an element of a group.
Ans. Let a be an element of a group G such that a2 = a.
To prove that a = e.
a2 = a a.a = a (aa) a–1 = aa–1
a(aa–1) = e
( aa–1 = e)
ae = e a = e
( ae = a)
2.7. If H is a subgroup of G such that x2 H for every x G, then
prove that H is a normal subgroup of G.
Ans. For any g G, h H; (gh)2 H and g–2 H.
Since H is a subgroup, h–1 g–2 H and so (gh)2 h–1 g–2 H. This
gives that gh gh h–1 g–2 H, i.e., ghg–1 H. Hence, H is a normal
subgroup of G.
Discrete Structures & Theory of Logic
SQ–9 C (CS/IT-Sem-3)
2.8. Prove that if a, b R then (a + b)2 = a2 + ab + ba + b2.
Ans. We have
(a + b)2 = (a + b) (a + b)
= a (a + b) + b (a + b) [by right distributive law]
= (aa + ab) + (ba + bb) [by left distributive law]
= a2 + ab + ba + b2.
2.9. In an integral domain D, show that if ab = ac with a 0
then b = c.
Ans. Since
ab = ac we have
ab – ac = 0 and so a (b – c) = 0
Since a 0, we must have b – c = 0, since D has no zero divisors.
Hence b = c.
2.10. Prove that left inverse of an element is also its right inverse
i.e.,
a–1 * a = e = a * a–1
Ans. Now a–1 * (a * a–1) = (a–1 * a)* a–1 (Associativity)
= e* a–1
= a–1 * e
–1
–1
Thus, a * (a * a ) = a–1 * e
a * a–1= e
Thus, the left inverse of an element in a group is also its right
inverse.
2.11. Define Lagrange’s theorem. What is the use of the theorem ?
Ans. Lagrange’s theorem :
Statement : The order of each subgroup of a finite group is a
divisor of the order of the group.
Use of theorem :
i. It can be used to find the subgroup of any order for the symmetric
group.
ii. It tells that the number of subgroups of the cyclic group of order p,
when p is prime then there is only one subgroup and that is {}.
SQ–10 C (CS/IT-Sem-3)
3
2 Marks Questions
Lattices and Boolean
Algebra
(2 Marks Questions)
3.1. Explain maximal and minimal element.
Ans. Maximal element : An element ‘a’ in the poset is called a maximal
element of P if a x for no ‘x’ in P, that is, if no element of P strictly
succeeds ‘a’.
Minimal element : An element ‘b’ in P is called a minimal element
of P if x b for no ‘x’ in P.
3.2. What do you mean by sublattice ?
Ans. A non-empty subset L of a lattice L is called a sublattice of L if
a, b L so that a b, a b L i.e., the algebra (L, , ) is a
sublattice of (L, , ) iff L is closed under both operations and.
3.3. Write the following in DNF (x + y) (x + y).
AKTU 2015-16, Marks 02
Ans. Given : (x + y) (x + y)
The complete CNF in two variables (x, y)
= (x + y) (x + y) (x + y) (x + y)
Hence, f (x, y) = (x + y) (x + y)
[f (x, y)] = [(x + y) (x + y)]
= xy + xy
which is the required DNF.
3.4. Explain lattice homomorphism and lattice isomorphism.
Ans. Lattice homomorphism : Let (L, *, ) and (S, , ) be two lattices.
A mapping g : L S is called a lattice homomorphism from the
lattice (L, *, ) to (S, , ) if for any a, b L
g(a * b) = g(a) g(b) and (g b) = g(a) g(b).
Lattice isomorphism : If a homomorphism g : L S of two
lattices (L, *, ) and (S, ,) is bijective i.e., one-to-one onto, then
g is called an isomorphism.
3.5. Define minterm and maxterm.
Ans. Minterm : A minterm of ‘n’ variables is a product of ‘n’ literals in
which each variable appears exactly once in either true or
complemented form, but not both.
SQ–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Maxterm : A maxterm of ‘n’ variables is a sum of ‘n’ literals in
which each variable appears exactly once in either true or
complemented form, but not both.
3.6. Show that the relation is a partial ordering on the set of
integers, Z.
Ans. Since :
1. a a for every a, is reflexive.
2. a b and b a imply a = b, is antisymmetric.
3. a b and b c imply a c, is transitive.
It follows that is a partial ordering on the set of integers and
(Z, ) is a poset.
3.7. Consider A = {x R : 1 < x < 2} with as the partial order.
Find
i. All the upper and lower bounds of A.
ii. Greatest lower bound and least upper bound of A.
Ans.
i. Every real number 2 is an upper bound of A and every real
number 1 is a lower bound of A.
ii. 1 is a greatest lower bound and 2 is the least upper bound of A.
3.8.
i.
ii.
iii.
Determine
All maximal and minimal elements
Greatest and least element
Upper and lower bounds of ‘a’ and ‘b’, ‘c’ and ‘d’
c
d
f
e
b
a
Fig. 3.8.1.
Ans.
i. Maximal elements = c, d, Minimal element = a, b
ii. Greatest and least elements do not exist.
iii. Upper bound for a,b are e, f, c, d.
Upper bound for c,d are does not exist.
Lower bound for a,b are does not exist.
Lower bound for c,d are f, e, a, b.
3.9. Let (A, ) be a distributive lattice. Show that if a x = a y
and a x = a y for some a then x = y.
Ans. We have
x = x (x a) = x (y a) ( Given condition)
= (x y) (x a)
SQ–12 C (CS/IT-Sem-3)
2 Marks Questions
=
=
x=
x=
(x y) (y a)
y (x a)
x (y a)
y
( Distributive property)
3.10. Prove that (A + B) (A + C) = A + BC
L.H.S = (A + B) (A + C)
= AA + AC + BA + BC
= A + AC + BA + BC
= A (1 + C) + BA + BC
= A + AB + BC
= A (1 + B) +BC = A + BC
= R.H.S.
Ans.
(
(
(
AA = 1)
1 + C =1)
1 + B =1)
3.11. For a given function, F = xy xy , find complement of F.
F = xy xy
F = xy
(A + A = A)
Take the complement of both sides,
Ans.
F = x y
Using De Morgan’s first law, we get,
F = xy
F = xy
3.12. Obtain an equivalent expression for [(x. y) (z + xy)].
Ans. Applying general form of De-Morgan’s theorem, we get
[(x.y) (z + xy)] = (x.y) + (z + xy)= x + y [z + (x + y)]
= x + y + zx + zy = x + x .z + y + yz = x + y + z
[Applying x + xy = x and x + xy = x + y]
3.13. Show that the “greater than or equal” relation (>=) is a
partial ordering on the set of integers.
AKTU 2016-17, Marks 02
Ans. Reflexive :
a a a Z (set of integer)
(a, a) A
R is reflexive.
Antisymmetric : Let (a, b) R and (b, a) R
a b and b a
a=b
R is antisymmetric.
Transitive : Let (a, b) R and (b, c) R
SQ–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
a b and b c
a c (a, c) R
R is transitive.
Hence, R is partial order relation.
3.14. Distinguish between bounded lattice and complemented
AKTU 2016-17, Marks 02
lattice.
Ans. Bounded lattice : A lattice which has both elements 0 and 1 is
called a bounded lattice.
Complemented lattice : A lattice L is called complemented lattice
if it is bounded and if every element in L has complement.
3.15. Draw the Hasse diagram of D30. AKTU 2018-19, Marks 02
Ans.
30
15
6
10
3
2
5
1
Fig. 3.15.1.
3.16. Define the terms : DNF and CNF.
Ans. Disjunction Normal Form (DNF) : A logical expression is said
to be in Disjunction Normal Form if it is the sum of elementary
product, i.e., join of elementary product.
Conjunctive Normal Form (CNF) : A logical expression is said
to be in Conjunctive Normal Form if it is the product of elementary
sums.
Example : p q, (p q) (~ p q)
SQ–14 C (CS/IT-Sem-3)
2 Marks Questions
4
Propositional Logic
and Predicate Logic
(2 Marks Questions)
4.1. What is compound proposition ?
Ans. A proposition obtained from the combinations of two or more
propositions by means of logical operators or connectives of two
or more propositions or by negating a single proposition is referred
to as composite or compound proposition.
4.2. Show the implications without constructing the truth
table (P Q) Q P Q.
AKTU 2016-17, Marks 02
Ans. (P Q) Q P Q
Take L.H.S
(P Q) Q =
=
=
=
(~P Q) Q
(~(~P Q)) Q
(P ~ Q) Q
(P Q) (~ Q Q)
= (P Q) T = P Q
It is equivalent.
4.3. Prove that (P Q) (P Q) is logically equivalent to P Q.
AKTU 2015-16, Marks 02
Ans. (P Q) (P Q) = P Q
P
Q
PQ
PQ
(P Q) (P Q)
PQ
T
T
T
T
T
T
T
F
T
F
F
F
F
T
T
F
F
F
F
F
F
F
T
T
4.4. The converse of a statement is : If a steel rod is stretched,
then it has been heated. Write the inverse of the statement.
AKTU 2015-16, Marks 02
Ans. The statement corresponding to the given converse is “If a steel
rod is stretched, then it has been heated”. Now the inverse of this
SQ–15 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
statement is “If a steel rod is not stretched then it has not been
heated”.
4.5. Give truth table for converse, contrapositive and inverse.
Ans. The truth table of the four propositions are as follows :
Conditional
Converse
Inverse
Contrapositive
p
q
pq
qp
~p ~q
~q~p
T
T
T
T
T
T
T
F
F
T
T
F
F
T
T
F
F
T
F
F
T
T
T
T
4.6. Show that contrapositive are logically equivalent; that is
~q~ppq
Ans. The truth table of ~ q ~ p and p q are shown below and the
logical equivalence is established by the last two columns of the
table, which are identical.
p
q
~p
~q
~q~p
pq
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
4.7. Give truth table for NOR and XOR.
Ans.
Truth table for NOR
p
q
T
T
T
F
F
p q
Truth table for XOR
p
q
p q
F
T
T
F
F
F
T
F
T
T
F
F
T
T
F
T
F
F
F
4.8. Verify that the proposition p (q ~p) is a contradiction.
SQ–16 C (CS/IT-Sem-3)
2 Marks Questions
Ans.
p
q
~p
q~p
p (q ~ p)
T
T
F
F
F
T
F
F
F
F
F
T
T
T
F
F
F
T
F
F
4.9. Show that [((p q) r) (~ p)] (q r) is tautology or
AKTU 2015-16, Marks 02
contradiction.
Ans.
p q
r p q ~p q r((p q) r)((pq) r) (~p)[((p q) r)
(~p)] (q r)
T T T
T
F
T
T
F
T
T T
F
T
F
F
F
F
T
T F T
T
F
F
T
F
T
T F
F
T
F
F
F
F
T
F T T
T
T
T
T
T
T
F T
F
T
T
F
F
F
T
F F T
F
T
F
T
T
F
F F
F
T
F
T
T
F
F
Question is incorrect. Since the result of the question is
contingency.
4.10. What are the contrapositive, converse, and the inverse of
the conditional statement : “The home team wins whenever
it is raining” ?
AKTU 2018-19, Marks 02
Ans. Given : The home team wins whenever it is raining.
q(conclusion) : The home team wins.
p(hypothesis) : It is raining.
Contrapositive : ~ q ~ p is “if the home team does not win
then it is not raining”.
Converse : q p is “if the home team wins then it is raining”.
Inverse : ~ p ~ q is “if it is not raining then the home team does
not win”.
Discrete Structures & Theory of Logic
SQ–17 C (CS/IT-Sem-3)
4.11. Show that (p q) and p q are logically equivalent.
AKTU 2018-19, Marks 02
Ans. To prove : (p + q) = p.q
To prove the theorem we will show that
(p + q) + p.q = 1
Consider (p + q) + p.q = {(p + q) + p}.{(p + q) + q}
by Distributive law
= {(q + p) + p}.{(p + q) + q}
by Commutative law
= {q + (p + p)}.{p + (q + q)}
by Associative law
= (q + 1).(p + 1)
by Complement law
= 1.1
by Dominance law
=1
...(4.11.1)
Also consider
(p + q).pq = pq.(p + q)
by Commutative law
= pq.p + pq.q
by Distributive law
= p.(pq) + p.(q q) by Commutative law
= (p. p).q + p(q. q)
by Associative law
= 0. q + p.0
by Complement law
= q.0 + p.0
by Commutative law
= 0+0
by Dominance law
=0
...(4.11.2)
From (4.11.1) and (4.11.2), we get,
pq is complement of (p + q) i.e., (p + q) = p q.
4.12. Find the contrapositive of “If he has courage, then he will
AKTU 2017-18, Marks 02
win”.
Ans. If he will not win then he does not have courage.
SQ–18 C (CS/IT-Sem-3)
5
2 Marks Questions
Trees, Graphs and
Combinatorics
(2 Marks Questions)
5.1. Explain trivial tree and non-trivial tree.
Ans. A tree with only one vertex is called a trivial tree. A tree with
more than one vertex is called non-trivial tree.
5.2. Describe the terms rank and nullity.
Ans. If ‘n’ be the number of vertices, ‘e’ be the number of edges and ‘k’
be (n k, k = 1 for connected graph) the number of components of
a graph G, then rank and nullity is defined as :
Rank, r = n – k
Nullity = e – n + k
5.3. Define binary tree traversal with example.
Ans. A traversal of tree is a process in which each vertex is visited
exactly once in a certain manner.
Binary tree traversal is of three types :
1. Preorder traversal
2. Postorder traversal
3. Inorder traversal
For example :
A
C
B
D
E
F
Fig. 5.3.1.
Preorder (NLR) : ABDECF
Postorder (LRN) : DEBFCA
Inorder (LNR) : DBEAFC
5.4. Define binary search trees.
Ans. A binary search tree is a binary tree T in which data is associated
with the vertices. The data are arranged so that, for each vertex V
in T, each data item in the left subtree of V is less than the data
SQ–19 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
item in V and each data item in the right subtree of V is greater
than the data item in V.
5.5. Give various representation of binary tree.
Ans. Various representation of binary tree are :
i. Storage representation
ii. Sequential representation
iii. Linked representation
5.6. Let G be a graph with ten vertices. If four vertices has degree
four and six vertices has degree five, then find the number
AKTU 2016-17, Marks 02
of edges of G.
Ans. We know that
deg (v ) = 2e
i
i
4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 5 = 2e
16 + 30 = 2e
2e = 46
e = 23
5.7. State the applications of binary search tree.
AKTU 2016-17, Marks 02
Ans. One of the most common applications is to efficiently store data in
sorted form in order to access and search stored elements quickly.
For example, std :: map or std :: set in C++ Standard Library.
Binary tree as data structure is useful for various implementations
of expression parsers and expression solvers.
5.8. Define multigraph. Explain with example in brief.
AKTU 2016-17, Marks 02
Ans. A multigraphs G(V, E) consists of a set of vertices V and a set of
edges E such that edge set E may contain multiple edges and self
loops.
Example :
a. Undirected multigraph :
v1
e1
v4
e2
v2
e6
e5
e3
e4
Fig. 5.8.1.
v3
SQ–20 C (CS/IT-Sem-3)
2 Marks Questions
b. Directed multigraph :
e3
v1
e1
v2
e2
e7
e5
e4
v4
v3
e6
Fig. 5.8.2.
5.9. Find the recurrence relation from yn = A2n + B(–3)n.
AKTU 2016-17, Marks 02
A2n
Ans. Given :
3)n
yn =
+ B (–
Therefore, yn + 1 = A(2)n + 1 + B (– 3)n + 1
= 2A (2)n – 3B (– 3)n
and
yn + 2 = A(2)n + 2 + B (– 3)n + 2
= 4A (2)n + 9B (– 3)n
Eliminating A and B from these equations, we get
yn
1
1
yn1 2 – 3 = 0
yn 2 4 9
= yn + 2 – yn + 1 – 6yn = 0 which is the
required recurrence relation.
5.10. Find the minimum number of students in a class to show
that five of them are born in same month.
Ans. Using pigeonhole principle :
Five of them are born in same month, so n = ?, m = 12.
n 1
1
5=
m
n1
12
48 = n – 1
n = 49
49 students are there to show that at least 5 of them are born in
same month.
4=
5.11. Explain edge coloring and k-egde coloring.
AKTU 2017-18, Marks 02
Ans. Edge coloring : An edge coloring of a graph G may also be thought
of as equivalent to a vertex coloring of the line graph L(G), the
graph that has a vertex for every edge of G and an edge for every
pair of adjacent edges in G.
Discrete Structures & Theory of Logic
SQ–21 C (CS/IT-Sem-3)
k-edge coloring : A proper edge coloring with k different colors is
called a (proper) k-edge coloring.
5.12. A tree has two vertices of degree 2, one vertex of degree 3
and three vertices of degree 4. How many vertices of degree
1 does it have ?
Ans. Let x be the required number. Now, total number of vertices
= 2+1+3+x=6+x
Hence the number of edges is 6 + x – 1 = 5 + x
[In a tree |E|= | V – 1|]
The total degree of the tree = 2 × 2 + 1 × 3 + 3 × 4 + 1 × x = 19 + x
So, the number of edges is
19 x
2
[2e = deg (v1)]
19 x
= 5+x
2
19 + x = 10 + 2x
or
x=9
Thus, there are 9 vertices of degree one in the tree.
Now,
5.13. State and prove pigeonhole principle.
AKTU 2015-16, Marks 02
Ans. Pigeonhole principle : If n pigeons are assigned to m pigeonholes
1.
2.
3.
4.
then at least one pigeon hole contains two or more pigeons (m < n).
Proof :
Let m pigeonholes be numbered with the numbers 1 through m.
Beginning with the pigeon 1, each pigeon is assigned in order to the
pigeonholes with the same number.
Since m < n i.e., the number of pigeonhole is less than the number
of pigeons, n – m pigeons are left without having assigned a pigeon
hole.
Thus, at least one pigeonhole will be assigned to a more than one
pigeon.
5.14. Draw the digraph G corresponding to adjacency matrix.
0 0 1 1
0 0 1 0
A=
1 1 0 1
1 1 1 0
Ans. Since the given matrix is square matrix of order four, the graph G
has 4 vertices v1, v2, v3 and v4. Draw an edge from vi to vj where
aij = 1. The required di graph is shown in Fig. 5.14.1.
SQ–22 C (CS/IT-Sem-3)
2 Marks Questions
v4
v2
v3
v1
Fig. 5.14.1.
5.15. A connected plane graph has 10 vertices each of degree 3.
Into how many regions, does a representation of this
planar graph split the plane ?
Ans. Here n = 10 and degree of each vertex is 3.
deg (v) = 3 × 10 = 30
But
deg (v) = 2e 30 = 2e e 15
By Euler’s formula, we have n – e + r = 2
10 – 15 + r = 2 r = 7.
5.16. How many permutations of the letters of the word
BANANA ?
Ans. There are 6 letters in the word BANANA of which three are alike
of one kind (3A’s), two are alike of second kind (2N’s) and the rest
one letter is different.
Hence, the required number of permutations =
6!
60 .
3!2!
5.17. How many 4 digit numbers can be formed by using the
digits 2, 4, 6, 8 when repetition of digits is allowed ?
AKTU 2015-16, Marks 02
Ans. When repetition is allowed :
The thousands place can be filled by 4 ways.
The hundreds place can be filled by 4 ways.
The tens place can be filled by 4 ways.
The units place can be filled by 4 ways.
Total number of 4 digit number = 4 × 4 × 4 × 4 = 256
5.18. How many bit strings of length eight either start with a ‘1’
bit or end with the two bit ‘00’ ?
AKTU 2018-19, Marks 02
Ans.
1. Number of bit strings of length eight that start with a 1 bit :
27 = 128.
2. Number of bit strings of length eight that end with bits 00 : 26 = 64.
Discrete Structures & Theory of Logic
SQ–23 C (CS/IT-Sem-3)
3. Number of bit strings of length eight 25 = 32 that start with a 1 bit
and end with bits 00 : 25 = 32
Hence, the number is 128 + 64 – 32 = 160.
5.19. Define Eulerian path, circuit and graph.
AKTU 2017-18, Marks 02
Ans. Eulerian path : A path of graph G which includes each edge of G
exactly once is called Eulerian path.
Eulerian circuit : A circuit of graph G which include each edge of
G exactly once.
Eulerain graph : A graph containing an Eulerian circuit is called
Eulerian graph.
5.20. Define chromatic number and isomorphic graph.
AKTU 2017-18, Marks 02
Ans. Chromatic number : The minimum number of colours required
for the proper colouring of a graph so that no two adjacent vertices
have the same colour, is called chromatic number of a graph.
Isomorphic graph : If two graphs are isomorphic to each other
then :
i. Both have same number of vertices and edges.
ii. Degree sequence of both graphs are same (degree sequence is the
sequence of degrees of the vertices of a graph arranged in nonincreasing order).
5.21. Define walk.
Ans. In a graph G, a finite alternating sequence of vertices and edges
1, e1, 2, e2 ... starting and ending with vertices such that each
edge in sequence is incident with the vertices following and
preceding, it is called walk.
5.22. Define non-planar graph.
Ans. A graph G is said to be non-planar graph if it cannot be drawn in
a plane so that no edges cross.
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B. Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2014-15
DISCRETE STRUCTURE AND
GRAPH THEORY
Time : 3 Hours
Max. Marks : 100
1. Attempt any four parts :
(5 × 4 = 20)
a. Show that R = {(a, b)|a b (mod m)} is an equivalence
relation on Z. Show that if x 1 y 1 and x 2 y 2 then
(x1 + x2) (y1 + y2).
b. Prove for any two sets A and B that, (A B)' = A' B'.
c. Let R be binary relation on the set of all strings of 0’s and
1’s such that R = {(a, b)|a and b are strings that have the
same number of 0’s}. Is R is an equivalence relation and a
partial ordering relation ?
d. If f : A B, g : B C are invertible functions, then show
that gof : A C is invertible and (gof)–1 = f–1 og–1.
e. Prove by the principle of mathematical induction, that
the sum of finite number of terms of a geometric
progression,
a + ar + ar2 + ... arn–1 = a(rn – 1)/(r – 1) if r 1.
f. Let A {1, 2, 3,...........,13}. Consider the equivalence relation
on A × A defined by (a, b) R (c, d) if a + d = b + c. Find
equivalence classes of (5,8).
2. Attempt any four parts :
(5 × 4 = 20)
a. Prove that (Z6, (+6)) is an abelian group of order 6, where
Z6 = {0, 1, 2, 3, 4, 5}.
b. Let G be a group and let a, b G be any elements. Then
i. (a–1)–1 = a
ii. (a * b)–1 = b–1 * a–1.
c. Prove that the intersection of two subgroups of a group is
also subgroup.
SP–2 C (CS/IT-Sem-3)
Solved Paper (2014-15)
d. Write and prove the Lagrange’s theorem. If a group
G = {...., – 3, 2, – 1, 0, 1, 2, 3,.....} having the addition as binary
operation. If H is a subgroup of group G where x2 H such
that x G. What is H and its left coset w.r.t 1 ?
e. Consider a ring (R, +, ) defined by a · a = a, determine
whether the ring is commutative or not.
f. Show that every group of order 3 is cyclic.
3. Attempt any two parts :
(10 × 2 = 20)
a. The directed graph G for a relation R on set A = {1, 2, 3, 4} is
shown below :
1
2
3
4
Fig. 1.
i. Verify that (A, R) is a poset and find its Hasse diagram.
ii. Is this a lattice ?
iii. How many more edges are needed in the Fig. 1 to extend (A,
R) to a total order ?
iv. What are the maximal and minimal elements ?
b. If the lattice is represented by the Hasse diagram given
below :
i. Find all the complements of ‘e’.
ii. Prove that the given lattice is bounded complemented
lattice.
b
a
c
f
d
e
Fig. 2.
c. Consider the Boolean function
a. f(x1, x2, x3, x4) = x1 + (x2. (x'1 + x4) + x3. (x'2 + x'4))
i. Simplify f algebraically.
Discrete Structures & Theory of Logic
SP–3 C (CS/IT-Sem-3)
ii. Draw the logic circuit of the f and the reduction of the f.
b. Write the expressions E1 = (x + xy) + (x/y) and E2 = x +
((xy + y)/y), into
i. Prefix notation ii. Postfix notation
4. Attempt any two parts :
(10 × 2 = 20)
a. i. Show that ((p q) ~ (~ p (~ q ~ r))) (~ p ~q ) (~p r)
is a tautology without using truth table.
ii. Rewrite the following arguments using quantifiers,
variables and predicate symbols :
a. All birds can fly.
b. Some men are genius.
c. Some numbers are not rational.
d. There is a student who likes mathematics but not
geography.
b. “If the labour market is perfect then the wages of all persons
in a particular employment will be equal. But it is always
the case that wages for such persons are not equal
therefore the labour market is not perfect”. Test the validity
of this argument using truth table.
c.
i.
ii.
iii.
iv.
v.
Explain the following terms with suitable example :
Conjunction
Disjunction
Conditional
Converse
Contrapositive
5. Attempt any two parts :
(10 × 2 = 20)
a. Solve the recurrence relation by the method of generating
function
ar – 7 ar–1 + 10ar–2 = 0, r 2
Given a0 = 3 and a1 = 3.
b. Solve the recurrence relation
ar+2 – 5ar+1 + 6 ar = (r + 1)2
c.
i.
ii.
iii.
Explain the following terms with example :
Homomorphism and Isomorphism graph
Euler graph and Hamiltonian graph
Planar and Complete bipartite graph
SP–4 C (CS/IT-Sem-3)
Solved Paper (2014-15)
SOLUTION OF PAPER (2014-15)
1. Attempt any four parts :
(5 × 4 = 20)
a. Show that R = {(a, b)|a b (mod m)} is an equivalence
relation on Z. Show that if x 1 y 1 and x 2 y 2 then
(x1 + x2) (y1 + y2).
Ans. R ={(a, b) | a b (mod m)}
For an equivalence relation it has to be reflexive, symmetric and
transitive.
Reflexive : For reflexive a we have (a, a) R i.e.,
a a (mod m)
a – a is divisible by m i.e., 0 is divisible by m
Therefore aRa, a Z, it is reflexive.
Symmetric : Let (a, b) Z and we have
(a, b) R i.e., a b (mod m)
a – b is divisible by m
a – b = km, k is an integer
(b – a) = (– k) m
(b – a) = p m, p is also an integer
b – a is also divisible by m
b a (mod m ) (b, a) R
It is symmetric.
Transitive : Let (a, b) R and (b, c) R then
(a , b) R a – b is divisible by m
a – b = t m, t is an integer
...(1)
(b, c) R b – c is divisible by m
b – c = s m, s is an integer
...(2)
From eq. (1) and (2)
a – b + b – c = (t + s) m
a – c = lm, l is also an integer
a – c is divisible by m
a c (mod m), yes it is transitive.
R is an equivalence relation.
To show : (x1 + x2) (y1 + y2) :
It is given x1 y1 and x2 y2
i.e., x1 – y1 divisible by m
x2 – y2 divisible by m
Adding above equation :
(x1 – y1) + (x2 – y2) is divisible by m
(x1 + x2) – (y1 + y2) is divisible by m
i.e., (x1 + x2) (y1 + y2)
b. Prove for any two sets A and B that, (A B)' = A' B'.
Ans.
Let
x (A B)
xAB
Discrete Structures & Theory of Logic
x A and x B
x A and x B
x A B
(A B) A B
Now, let
x A B
x A and x B
x A and x B
x (A B)
x (A B)
(AB) (A B)
From eq. (1) and (2), (A B) = A B
SP–5 C (CS/IT-Sem-3)
...(1)
...(2)
c. Let R be binary relation on the set of all strings of 0’s and
1’s such that R = {(a, b)|a and b are strings that have the
same number of 0’s}. Is R is an equivalence relation and a
partial ordering relation ?
Ans. For equivalence relation :
Reflexive : a R a (a, a) R a R
where a is a string of 0’s and 1’s.
Always a is related to a because both a has same number of 0’s.
It is reflexive.
Symmetric : Let (a, b) R
then a and b both have same number of 0’s which indicates that
again both b and a will also have same number of zeros. Hence (b,
a) R. It is symmetric.
Transitive : Let (a, b) R, (b, c) R
(a, b) R a and b have same number of zeros.
(b, c) R b and c have same number of zeros.
Therefore a and c also have same number of zeros, hence (a, c) R.
It is transitive.
R is an equivalence relation.
For partial order, it has to be reflexive, antisymmetric and transitive.
Since, symmetricity and antisymmetricity cannot hold together.
Therefore, it is not partial order relation.
d. If f : A B, g : B C are invertible functions, then show
that g f : A C is invertible and (g f)–1 = f–1 g–1.
Ans. If f: A B and g : B C be one-to-one onto functions, then g f is
also one-onto and (g f)–1 = f–1 g–1
Proof. Since f is one-to-one, f (x1) = f (x2) x1 = x2 for x1, x2 R
Again since g is one-to-one, g (y1) = g (y2) y1 = y2 for y1, y2 R
Now g f is one-to-one, since (g f) (x1) = (g f) (x2) g [f(x1)] = g
[f (x2)]
f (x1) = f(x2 )
[g is one-to-one]
x1 = x2
[f is one-to-one]
Since g is onto, for z C, there exists y B such that g (y) = z. Also
f being onto there exists x A such that f (x) = y. Hence z = g (y)
SP–6 C (CS/IT-Sem-3)
Solved Paper (2014-15)
= g [f (x)] = (g f) (x)
This shows that every element z C has pre-image under g f. So,
g f is onto.
Thus, g f is one-to-one onto function and hence (g f)–1 exists.
By the definition of the composite functions, g f : A C. So,
(g f)–1 : C A.
Also g–1 : C B and f–1 : B A.
Then by the definition of composite functions, f–1 g–1 : C A.
Therefore, the domain of (g f)–1 = the domain of f–1 g–1.
Now(g f)–1 (z) = x (g f) (x) = z
g (f (x)) = z
g (y) = z where y = f (x)
y = g–1 (z)
f–1 (y) = f–1 (g–1 (z)) = (f–1 g–1) (z)
x = (f–1 g–1) (z)
[f–1 (y) = x]
Thus, (g f)–1 (z) = (f–1 g–1) (z).
So,
(g f)–1 = f–1 g–1.
e. Prove by the principle of mathematical induction, that
the sum of finite number of terms of a geometric
progression,
a + ar + ar2 + ... arn–1 = a(rn – 1)/(r – 1) if r 1.
Ans. Basis : True for n = 1 i.e.,
L.H.S = a
a (r 1)
=a
r1
Therefore, L.H.S. = R.H.S.
Induction : Let it be true for n = k i.e.,
R.H.S =
a (r k 1)
r 1
Now we will show that it is true for n = k + 1 using eq. (1)
i.e., a + ar + ar2 +.... + ark-1 + ark
Using eq. (1), we get
a + ar + ar2 + .... + ark–1 =
...(1)
a (r k 1)
+ ark
r 1
ar k a ar k1 ar k
a ( r k 1 1)
=
r 1
r 1
which is R.H.S. for n = k + 1, hence it is true for n = k + 1.
By mathematical induction, it is true for all n.
=
f. Let A {1, 2, 3,...........,13}. Consider the equivalence relation
on A × A defined by (a, b) R (c, d) if a + d = b + c. Find
equivalence classes of (5,8).
A = {1, 2, 3, ...., 13}
Ans.
SP–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
[(5, 8)] =
=
=
[5, 8] =
[(a, b) : (a, b) R (5, 8), (a, b) A × A]
[(a, b) : a + 8 = b + 5]
[(a, b) : a + 3 = b]
{(1, 4), (2, 5), (3, 6), (4, 7)
(5, 8), (6, 9), (7, 10), (8, 11)
(9, 12), (10, 13)}
2. Attempt any four parts :
(5 × 4 = 20)
a. Prove that (Z6, (+6)) is an abelian group of order 6, where
Z6 = {0, 1, 2, 3, 4, 5}.
Ans. The composition table is :
+6
0
1
2
3
4 5
0
1
2
3
4
5
0
1
2
3
4
5
1
2
3
4
5
0
2
3
4
5
0
1
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
Since
2 +6 1 = 3
4 +6 5 = 3
From the table we get the following observations :
Closure : Since all the entries in the table belong to the given set
Z6. Therefore, Z6 is closed with respect to addition modulo 6.
Associativity : The composition ‘+6’ is associative. If a, b, c are any
three elements of Z6,
a +6 (b +6 c) =a +6 (b + c)
[ b +6 c = b + c (mod 6)]
= least non-negative remainder when a + (b + c) is divided by 6.
= least non-negative remainder when (a + b) + c is divided by 6.
= (a + b) +6 c = (a +6 b) +6 c.
Identity : We have 0 Z6. If a is any element of Z6, then from the
composition table we see that
0 +6 a = a = a +6 0
Therefore, 0 is the identity element.
Inverse : From the table we see that the inverse of 0, 1, 2, 3, 4, 5
are 0, 5, 4, 3, 2, 1 respectively. For example 4 +6 2 = 0 = 2 +6 4 implies
4 is the inverse of 2.
Commutative : The composition is commutative as the elements
are symmetrically arranged about the main diagonal. The number
of elements in the set Z6 is 6.
(Z6, +6) is a finite abelian group of order 6.
b. Let G be a group and let a, b G be any elements. Then
i. (a–1)–1 = a
ii. (a * b)–1 = b–1 * a–1.
Ans.
i. Let e be the identity element for * in G.
SP–8 C (CS/IT-Sem-3)
Solved Paper (2014-15)
Then we have a * a – 1 = e, where a – 1 G.
Also (a – 1) – 1 * a – 1 = e
Therefore, (a – 1) – 1 * a – 1 = a * a – 1.
Thus, by right cancellation law, we have (a – 1) – 1 = a.
ii. Let a and b G and G is a group for *, then a * b G (closure)
Therefore, (a * b) – 1 * (a * b) = e.
....(1)
Let a – 1 and b – 1 be the inverses of a and b respectively, then a – 1,
b – 1 G.
Therefore, (b – 1 * a – 1) * (a * b) = b – 1 * (a – 1 * a) * b (associativity)
= b–1 * e * b = b–1 * b = e
...(2)
From eq. (1) and (2) we have,
(a * b) – 1 * (a * b) = (b – 1 * a – 1)* (a * b)
(a * b) – 1 = b – 1 * a –1
(by right cancellation law)
c. Prove that the intersection of two subgroups of a group is
also subgroup.
Ans. Let H1 and H2 be any two subgroups of G. Since at least the identity
element e is common to both H1 and H2.
H1 H2
In order to prove that H1 H2 is a subgroup, it is sufficient to prove
that
a H1 H2, b H1 H2 ab–1 H1 H2
Now a H1 H2 a H1 and a H2
b H1 H2 b H1 and b H2
But H1, H2 are subgroups. Therefore,
a H1, b H1 ab–1 H1
a H2 , bH2 ab–1 H2
Finally, ab–1 H1, ab–1 H2 ab–1 H1 H2
Thus, we have shown that
a H1 H2, b H1 H2 ab–1 H1 H2.
Hence, H1 H2 is a subgroup of G.
d. Write and prove the Lagrange’s theorem. If a group
G = {...., – 3, 2, – 1, 0, 1, 2, 3,.....} having the addition as binary
operation. If H is a subgroup of group G where x2 H such
that x G. What is H and its left coset w.r.t 1 ?
Ans. Lagrange’s theorem :
If G is a finite group and H is a subgroup of G then o(H) divides
o(G). Moreover, the number of distinct left (right) cosets of H in G
is o(G)/o(H).
Proof : Let H be subgroup of order m of a finite group G of order n.
Let H {h1, h2, ..., hm}
Let a G. Then aH is a left coset of H in G and aH = {ah1, ah2, ...,
ahm} has m distinct elements as ahi = ahj hi = hj by cancellation
law in G.
Thus, every left coset of H in G has m distinct elements.
Discrete Structures & Theory of Logic
SP–9 C (CS/IT-Sem-3)
Since G is a finite group, the number of distinct left cosets will also
be finite. Let it be k. Then the union of these k-left cosets of H in G
is equal to G.
i.e., if a1H, a2H, ..., akH are right cosets of H in G then
G = a1H a2H ... akH.
o(G) = o(a1H) + o(a2H) + ... + o(akH)
(Since two distinct left cosets are mutually disjoint.)
n = m + m + ... + m (k times)
n
n = mk k =
m
o(G )
k=
.
o( H )
Thus order of each subgroup of a finite group G is a divisor of the
order of the group.
Numerical :
H = {x2 : x G} = {0, 1, 4, 9, 16, 25 ....}
Left coset of H will be 1 + H = {1, 2, 5, 10, 17, 26, ....}
e. Consider a ring (R, +, ) defined by a · a = a, determine
whether the ring is commutative or not.
Ans. Let
a, b R (a + b)2 = (a + b)
(a + b) (a + b) = (a + b)
(a + b)a + (a + b)b = (a + b)
(a2 + ba) + (ab + b2) = (a + b)
(a + ba) + (ab + b) = (a + b)
( a2 = a and b2 = b)
(a + b) + (ba + ab) = (a + b) + 0
ba + ab = 0
a + b = 0 a + b = a + a [being every element of its own additive
inverse]
b=a
ab = ba
R is commutative ring.
f. Show that every group of order 3 is cyclic.
Ans.
1. Suppose G is a finite group whose order is a prime number p, then
to prove that G is a cyclic group.
2. An integer p is said to be a prime number if p 0, p 1, and if the
only divisors of p are ± 1, ± p.
3. Some G is a group of prime order, therefore G must contain at least
2 element. Note that 2 is the least positive prime integer.
4. Therefore, there must exist an element a G such that a the
identity element e.
SP–10 C (CS/IT-Sem-3)
Solved Paper (2014-15)
5. Since a is not the identity element, therefore o(a) is definitely 2.
Let o(a) = m. If H is the cyclic subgroup of G generated by a then
o (H = o (a) = m).
6. By Lagrange’s theorem m must be a divisor of p. But p is prime and
m 2. Hence, m = p.
7. H = G. Since H is cyclic therefore G is cyclic and a is a generator
of G.
3. Attempt any two parts :
(10 × 2 = 20)
a. The directed graph G for a relation R on set A = {1, 2, 3, 4} is
shown below :
1
2
3
4
Fig. 1.
i. Verify that (A, R) is a poset and find its Hasse diagram.
ii. Is this a lattice ?
iii. How many more edges are needed in the Fig. 1 to extend (A,
R) to a total order ?
iv. What are the maximal and minimal elements ?
Ans.
i. The relation R corresponding to the given directed graph is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 1), (3, 4), (1, 4), (3, 2)}
R is a partial order relation if it is reflexive, antisymmetric and
transitive.
Reflexive : Since aRa, a A . Hence, it is reflexive.
Antisymmetric : Since aRb and bRa then we get a = b otherwise
aRb or bRa.
Hence, it is antisymmetric.
Transitive: For every aRb and bRc we get aRc. Hence, it is
transitive.
Therefore, we can say that (A, R) is poset. Its Hasse diagram is :
1
2
4
3
Fig. 2.
SP–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
ii. Since there is no lub of 1 and 2 and same for 2 and 4. The given
poset is not a lattice.
1 2 3 4
1 1 1 1
2 2 2
3 1 2 3 1
4 1 1 4
iii. Only one edge (4, 2) is included to make it total order.
iv. Maximals are {1, 2} and minimals are {3, 4}.
b. If the lattice is represented by the Hasse diagram given
below :
i. Find all the complements of ‘e’.
ii. Prove that the given lattice is bounded complemented
lattice.
b
c
a
f
d
e
Fig. 3.
Ans.
i. In a given lattice, greatest element is b and least element is e. An
element x in lattice is called a complement of element y if
y x = b and y x = e
For element e,
eb=b , e b=e
So, complement of e is b
ii. Proof :
For bounded complemented lattice, every element in lattice has a
complement and lattice is bounded. Since, given lattice have greatest
and least element. So, the given lattice is bounded.
Now the complement of all elements is given below :
Complement of a = {c, d}
Complement of b = {e}
Complement of c = {a, f}
Complement of d = {a, f}
Complement of e = {b}
Complement of f = {c, d}
Since, complement of every element exists and lattice is bounded.
So, the given lattice is bounded complemented lattice.
Hence proved.
SP–12 C (CS/IT-Sem-3)
Solved Paper (2014-15)
c.
a.
i.
ii.
b.
Consider the Boolean function
f(x1, x2, x3, x4) = x1 + (x2. (x'1 + x4) + x3. (x'2 + x'4))
Simplify f algebraically.
Draw the logic circuit of the f and the reduction of the f.
Write the expressions E1 = (x + xy) + (x/y) and E2 = x +
((xy + y)/y), into
i. Prefix notation ii. Postfix notation
Ans.
a. i.
f(x1, x2, x3, x4) = x1 + (x2.(x1 + x4) + x3.(x2+ x4)
= x1 + x2.x1 + x2.x4 + x3.x2 + x3.x4
= x1 + x2 + x2.x4 + x3.x2 + x3.x4
= x1 + x2.(1 + x4) + x3.x2 + x3.x4
= x1 + x2 + x3.x2 + x3.x4
= x1 + x2 + x3 + x3.x4
= x1 + x2 + x3.(1 + x4)
ii. Logic circuit :
= x1 + x2 + x3
x1
x2
x3
x1 + x2 + x3
Fig. 4.
Reduction of f :
f (x1, x2, x3, x4) = x1 + (x2.(x1 + x4) + x3.(x2 + x4)
= x1 + (x2.x1 + x2.x4) + (x3.x2 + x3.x4)
= x1 + x2.x1 + x2.x4 + x3.x2 + x3.x4
b. E1 = (x + x * y) + (x/y)
Binary tree is :
+
+
x
/
x
*
x
y
Fig. 5.
Prefix : ++ x * x y / xy
Postfix : xx y * + xy / +
E2 = x + ((x * y + y)/y)
y
SP–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Binary tree is :
+
/
x
y
+
y
*
y
x
Fig. 6.
Prefix : + x / + * x y y y
Postfix : x x y * y + y / +
4. Attempt any two parts :
(10 × 2 = 20)
a. i. Show that ((p q) ~ (~ p (~ q ~ r))) (~ p ~q ) (~p r)
is a tautology without using truth table.
ii. Rewrite the following arguments using quantifiers,
variables and predicate symbols :
a. All birds can fly.
b. Some men are genius.
c. Some numbers are not rational.
d. There is a student who likes mathematics but not
geography.
Ans.
i. We have
((p q) ~ (~ p (~ q r))) (~ p ~ q) (~ p r)
((p q) ~ (~ p ~ (q r))) (~ (p q) ~ ( p r))
(Using De Morgan's Law)
[(p q)] (p (q r)) ~ ((p q) (p r))
[(p q) (p q) (p r)] ~ (( p q) (p r))
(Using Distributive Law)
[((p q) (p q)] (p r) ~ ((p q) (p r))
((p q) (p r)) ~ ((p q) (p r))
x ~ x where x = (p q) (p r)
T
ii.
a. x [B(x) F(x)]
b. x [M(x) G(x)]
c. ~[ (x) (N(x) R(x)]
d. x [S(x) M(x) ~ G(x)]
b. “If the labour market is perfect then the wages of all persons
in a particular employment will be equal. But it is always
the case that wages for such persons are not equal
therefore the labour market is not perfect”. Test the validity
of this argument using truth table.
SP–14 C (CS/IT-Sem-3)
Solved Paper (2014-15)
Ans. Let p1 : The labour market is perfect.
p2 : Wages of all persons in a particular employment will be equal.
~p2 : Wages for such persons are not equal.
~p1 : The labour market is not perfect.
The premises are p1 p2 , ~ p2 and the conclusion is ~ p1. The argument
p1 p2, ~ p2 ~ p1 is valid if ((p1 p2) ~ p2) ~ p1 is a tautology.
Its truth table is,
~p2 p1 p2 (p1 p2) ~p2 (p1p2 ~ p2) ~p1
p1
p2
~p1
T
T
F
F
T
F
T
T
F
F
T
F
F
T
F
T
T
F
T
F
T
F
F
T
T
T
T
T
Since ((p1 p2 ) ~p2) ~p1 is a tautology. Hence, this is valid
argument.
c.
i.
ii.
iii.
iv.
v.
Explain the following terms with suitable example :
Conjunction
Disjunction
Conditional
Converse
Contrapositive
Ans.
i. Conjunction : If p and q are two statements, then conjunction of
p and q is the compound statement denoted by p q and read as
“p and q”. Its truth table is,
p
q
pq
T
T
T
T
F
F
F
T
F
F
F
F
Example :
p : Ram is healthy.
q : He has blue eyes.
p q : Ram is healthy and he has blue eyes.
ii. Disjunction : If p and q are two statements, the disjunction of
p and q is the compound statement denoted by p q and it is read as
“p or q”. Its truth table is,
Discrete Structures & Theory of Logic
p
q
pq
T
T
T
T
F
T
F
T
T
F
F
F
SP–15 C (CS/IT-Sem-3)
Example :
p : Ram will go to Delhi.
q : Ram will go to Calcutta.
p q : Ram will go to Delhi or Calcutta.
iii. Conditional : If p and q are propositions. The compound proposition
if p then q denoted by p q or p q and is called conditional
proposition or implication. It is read as “If p then q” and its truth
table is,
p
q
pq
T
T
T
T
F
F
F
T
T
F
F
T
Example :
p : Ram works hard.
q : He will get good marks.
p q : If Ram works hard then he will get good marks.
For converse and contrapositive :
Let
p : It rains.
q : The crops will grow.
iv. Converse : If p q is an implication then its converse is given by
q p states that S : If the crops grow, then there has been rain.
v. Contrapositive : If p q is an implication then its contrapositive
is given by q p states that,
t : If the crops do not grow then there has been no rain.
Inverse :
If p q is implication the inverse of p q is ~ p ~ q.
Consider the statement
p : It rains.
q : The crops will grow
The implication p q states that,
r : If it rains then the crops will grow.
The inverse of the implication p q, namely ~ p ~ q states that.
u : If it does not rain then the crops will not grow.
SP–16 C (CS/IT-Sem-3)
Solved Paper (2014-15)
5. Attempt any two parts :
(10 × 2 = 20)
a. Solve the recurrence relation by the method of generating
function
ar – 7 ar – 1 + 10ar – 2 = 0, r 2
Given a0 = 3 and a1 = 3.
Ans. ar – 7ar – 1 + 10 ar – 2 = 0, r 2
Multiply by xr and take sum from 2 to .
r2
r2
r2
ar xr 7 ar –1 xr 10 ar 2 xr 0
(a2 x2 + a3 x3 + a4 x4 + ....) – 7 (a1 x2 + a2 x3 + ....)
+ 10 (a0 x2 + a1 x3 + ....) = 0
We know that
G(x) =
ar xr a0 a1 x ....
r0
G(x) – a0 – a1 x – 7 x (G(x) – a0) + 10 x2 G(x) = 0
G(x) [1 – 7 x + 10 x2] = a0 + a1 x – 7 a0 x
= 3 + 3x – 21 x = 3 – 18 x
G(x) =
=
3 18 x
3 18 x
10 x 2 7 x 1 10 x 2 5 x 2 x 1
3 18 x
3 18 x
5 x (2 x 1) 1 (2 x 1) (5 x – 1)(2 x 1)
Now,
3 18 x
A
B
=
(5 x 1) (2 x 1)
5 x 1 2 x 1
3 – 18 x = A (2 x – 1) + B (5 x – 1)
put
x=
1
2
5
3
3 – 9 = B 1 – 6 =
B B = – 4
2
2
put
x=
1
5
3
3
2
18
= A 1 A 1 A = 1
5
5
5
5
1
4
4
1
G(x) =
=
5 x 1 2x 1
1 2x 1 5x
ar = 4.2r – 5r
3–
b. Solve the recurrence relation
ar+2 – 5ar+1 + 6 ar = (r + 1)2
Ans. ar+2 – 5 ar+1 + 6 ar = (r + 1)2 = r2 + 2r + 1
Now the characteristic equation is :
...(1)
SP–17 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
x2 – 5 x + 6 = 0
(x – 3) (x – 2) = 0 x = 3, 2
The homogeneous solution is :
ar(h) = C12r + C2 3r
Let the particular solution be :
ar(p) = A0 + A1r + A2r2
From eq. (1)
A0 + A1 (r + 2) + A2 (r + 2)2 – 5 {A0 + A1 (r + 1)} + A2 (r + 1)2}
+ 6A0 + 6A1r + 6A2 r2
=
r2
+2r+1
(A0 + 2A1 + 4A2 –5A0 – 5A1 – 5A2 + 6A0) + r (A1 + 4A2 – 5A1 – 10A2 +
6A1)
+ r2 ( A2 – 5A2 + 6A2) = r2 + 2r + 1
Comparing both sides, we get,
2A0 – 3A1 – A2 = 1
...(2)
2A1 – 6A2 = 2
2A2 = 1
...(3)
A2 = 1/2
From eq. (3), 2A1 – 3 = 2
A1 =
5
2
From eq. (2)
2A0 –
15 1
=1
2 2
2A0 – 8 = 1
A0 =
9
2
9 5
r2
r
2 2
2
The final solution is,
ar(p) =
ar = ar(h) + ar(p) = C1 2r + C2 3r +
c.
i.
ii.
iii.
9
5
r2
+ r
2
2
2
Explain the following terms with example :
Homomorphism and Isomorphism graph
Euler graph and Hamiltonian graph
Planar and Complete bipartite graph
Ans.
i. Homomorphism of graph : Two graphs are said to be
homomorphic if one graph can be obtained from the other by the
creation of edges in series (i.e., by insertion of vertices of degree
two) or by the merger of edges in series.
SP–18 C (CS/IT-Sem-3)
Solved Paper (2014-15)
(a)
(b )
Fig. 7.
(c )
Isomorphism of graph : Two graphs are isomorphic to each
other if :
i. Both have same number of vertices and edges.
ii. Degree sequence of both graphs are same (degree sequence is the
sequence of degrees of the vertices of a graph arranged in nonincreasing order).
Example :
Fig. 8.
ii. Eulerian path : A path of graph G which includes each edge of G
exactly once is called Eulerian path.
Eulerian circuit : A circuit of graph G which include each edge of
G exactly once.
Eulerain graph : A graph containing an Eulerian circuit is called
Eulerian graph.
For example : Graphs given below are Eulerian graphs.
V1
V3
A
B
V4 D
Fig. 9.
C
V2
V5
Hamiltonian graph : A Hamiltonian circuit in a graph G is a
closed path that visit every vertex in G exactly once except the end
vertices. A graph G is called Hamiltonian graph if it contains a
Hamiltonian circuit.
For example : Consider graphs given below :
C
D
C
D
F
A
B
A
(a)
E
B
(b )
Fig. 10.
Graph given is Fig. 10(a) is a Hamiltonian graph since it contains a
Hamiltonian circuit A – B – C – D – A while graph in Fig 10(b) is not
a Hamiltonian graph.
Hamiltonian path : The path obtained by removing any one edge
from a Hamiltonian circuit is called Hamiltonian path. Hamiltonian
path is subgraph of Hamiltonian circuit. But converse is not true.
Discrete Structures & Theory of Logic
SP–19 C (CS/IT-Sem-3)
The length of Hamiltonian path in a connected graph of n vertices
is n – 1 if it exists.
iii. Planar graph :
A graph G is said to be planar if there exists some geometric
representation of G which can be drawn on a plane such that no
two of its edges intersect except only at the common vertex.
i. A graph is said a planar graph, if it cannot be drawn on a plane
without a crossover between its edges crossing.
ii. The graphs shown in Fig. 11(a) and (b) are planar graphs.
(a)
Fig. 11. Some planar graph.
(b )
Complete bipartite graph : The complete bipartite graph on m
and n vertices, denoted Km, n is the graph, whose vertex set is
partitioned into sets V1 with m vertices and V2 with n vertices in
which there is an edge between each pair of vertices 1 and 2
where 1 is in V1 and 2 is in V2. The complete bipartite graphs
K2,3, K2,4, K 3,3, K3,5, and K2,6
K2,3
K2,4
K3,5
K3,3
K2,6
Fig. 12. Some complete bipartite graphs.
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B.Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2015-16
DISCRETE STRUCTURE AND
GRAPH THEORY
Time : 3 Hours
Max. Marks : 100
Note : Attempt all parts. All parts carry equal marks. Write answer of
each part in short.
(2 × 10 = 20)
SECTION – A
1. a. Define multiset and power set. Determine the power set
A = {1, 2}.
b. Show that [((p q) r) (~ p)] (q r) is tautology or
contradiction.
c. State and prove pigeonhole principle.
d. Show that if set A has 3 elements, then we can have 26
symmetric relation on A.
e. Prove that (P Q) (P Q) is logically equivalent to P Q.
f. How many 4 digit numbers can be formed by using the
digits 2, 4, 6, 8 when repetition of digits is allowed ?
g. The converse of a statement is : If a steel rod is stretched,
then it has been heated. Write the inverse of the statement.
h. If a and b are any two elements of group G then prove
(a * b) – 1 = (b – 1 * a – 1).
i. If f : A B is one-one onto mapping, then prove that
f –1 : B A will be one-one onto mapping.
j. Write the following in DNF (x + y) (x + y).
SP–2 C (CS/IT-Sem-3)
Solved Paper (2015-16)
SECTION – B
2. Prove that n3 + 2n is divisible by 3 using principle of
mathematical induction, where n is natural number.
3. Solve the recurrence relation using generating function :
an – 7an – 1 + 10an – 2 = 0 with a0 = 3, a1 = 3.
4. Express the following statements using quantifiers and
logical connectives.
a. Mathematics book that is published in India has a blue
cover.
b. All animals are mortal. All human being are animal.
Therefore, all human being are mortal.
c. There exists a mathematics book with a cover that is not
blue.
d. He eats crackers only if he drinks milk.
e. There are mathematics books that are published outside
India.
f. Not all books have bibliographies.
5. Draw the Hasse diagram of [P (a, b, c), ] (Note : ‘’ stands
for subset). Find greatest element, least element, minimal
element and maximal element.
6. Simplify the following boolean expressions using k-map :
a. Y = ((AB) + A + AB)
b. A B C D + A B C D + A B CD + A B CD = A B
7. Let G be the set of all non-zero real number and let
a*b = ab/2. Show that (G*) be an abelian group.
8. The following relation on A = {1, 2, 3, 4}. Determine whether
the following :
a. R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}.
b. R = AXA
Is an equivalence relation or not.
9. If the permutation of the elements of {1, 2, 3, 4, 5} are given
by a = (1 2 3) (4 5), b = (1) (2) (3) (4 5), c = (1 5 2 4) (3). Find the
value of x, if ax = b. And also prove that the set Z4 = (0, 1, 2, 3)
is a commutative ring with respect to the binary modulo
operation +4 and *4.
Discrete Structures & Theory of Logic
SP–3 C (CS/IT-Sem-3)
SECTION – C
10. Let L be a bounded distributed lattice, prove if a complement
exists, it is unique. Is D12 a complemented lattice ? Draw
the Hasse diagram of [P (a, b, c), <], (Note : ‘<’ stands for
subset). Find greatest element, least element, minimal
element and maximal element.
11. Determine whether each of these functions is a bijective
from R to R.
a. f(x) = x2 + 1
b. f(x) = x3
c. f(x) = (x2 + 1)/(x2 + 2)
12. a. Prove that inverse of each element in a group is unique.
b. Show that G = [(1, 2, 4, 5, 7, 8), ×9] is cyclic. How many
generators are there ? What are they ?
SP–4 C (CS/IT-Sem-3)
Solved Paper (2015-16)
SOLUTION OF PAPER (2015-16)
Note : Attempt all parts. All parts carry equal marks. Write answer of
each part in short.
(2 × 10 = 20)
SECTION – A
1. a. Define multiset and power set. Determine the power set
A = {1, 2}.
Ans. Multiset : Multisets are sets where an element can occur as a
member more than once.
For example : A = { p, p, p, q, q, q, r, r, r, r}
B = { p, p, q, q, q, r}
are multisets.
Power set : A power set is a set of all subsets of the set.
The power set of A = {1, 2} is {{}, {1}, {2}}.
b. Show that [((p q) r) (~ p)] (q r) is tautology or
contradiction.
Ans.
p q
r p q ~p q r((p q) r)((pq) r) (~p)[((p q) r)
(~p)] (q r)
T T T
T
F
T
T
F
T
T T
F
T
F
F
F
F
T
T F T
T
F
F
T
F
T
T F
F
T
F
F
F
F
T
F T T
T
T
T
T
T
T
F T
F
T
T
F
F
F
T
F F T
F
T
F
T
T
F
F F
F
T
F
T
T
F
F
Question is incorrect. Since the result of the question is
contingency.
c. State and prove pigeonhole principle.
Ans. Pigeonhole principle : If n pigeons are assigned to m pigeonholes
then at least one pigeon hole contains two or more pigeons (m < n).
Proof :
1. Let m pigeonholes be numbered with the numbers 1 through m.
2. Beginning with the pigeon 1, each pigeon is assigned in order to the
pigeonholes with the same number.
Discrete Structures & Theory of Logic
SP–5 C (CS/IT-Sem-3)
3. Since m < n i.e., the number of pigeonhole is less than the number
of pigeons, n – m pigeons are left without having assigned a pigeon
hole.
4. Thus, at least one pigeonhole will be assigned to a more than one
pigeon.
d. Show that if set A has 3 elements, then we can have 26
symmetric relation on A.
Ans. Number of elements in set = 3
Number of symmetric relations if number of elements is
n = 2n(n + 1)/2
Here,
n=3
Number of symmetric relations
= 23(3 + 1)/2
= 23(4)/2
= 26
Hence proved.
e. Prove that (P Q) (P Q) is logically equivalent to P Q.
Ans. (P Q) (P Q) = P Q
P
Q
PQ
PQ
(P Q) (P Q)
PQ
T
T
T
T
T
T
T
F
T
F
F
F
F
T
T
F
F
F
F
F
F
F
T
T
f. How many 4 digit numbers can be formed by using the
digits 2, 4, 6, 8 when repetition of digits is allowed ?
Ans. When repetition is allowed :
The thousands place can be filled by 4 ways.
The hundreds place can be filled by 4 ways.
The tens place can be filled by 4 ways.
The units place can be filled by 4 ways.
Total number of 4 digit number = 4 × 4 × 4 × 4 = 256
g. The converse of a statement is : If a steel rod is stretched,
then it has been heated. Write the inverse of the statement.
Ans. The statement corresponding to the given converse is “If a steel
rod is stretched, then it has been heated”. Now the inverse of this
statement is “If a steel rod is not stretched then it has not been
heated”.
SP–6 C (CS/IT-Sem-3)
Solved Paper (2015-16)
h. If a and b are any two elements of group G then prove
(a * b) – 1 = (b – 1 * a – 1).
Ans. Consider (a * b) * (b – 1 * a – 1)
= a * (b * b – 1) * a – 1
= a * e * a– 1
= a * a–1= e
Also (b – 1 * a – 1) * (a * b) = b – 1 * (a – 1 * a) * b
= b–1 * e * b
= b–1 * b = e
Therefore (a * b) – 1 = b – 1 * a – 1 for any a, b G
i. If f : A B is one-one onto mapping, then prove that
f –1 : B A will be one-one onto mapping.
Ans. Proof : Here f : A B is one-to-one and onto.
a1, a2 A and b1, b2 B so that
b1 = f(a1), b2 = f (a2) and a1 = f–1(b1), a2 = f–1 (b2)
As f is one-to-one
f(a1) = f(a2) a1 = a2
b1 = b2 f–1(b1) = f–1(b2)
–1
i.e.,
f (b1) = f–1(b2) b1 = b2
f–1 is one-to-one function.
As f is onto.
Every element of B is associated with a unique element of A i.e.,
for any a A is pre-image of some b B where b = f(a) a = f–1(b)
i.e., for b B, there exists f–1 image a A.
Hence, f–1 is onto.
j. Write the following in DNF (x + y) (x + y).
Ans. Given : (x + y) (x + y)
The complete CNF in two variables (x, y)
= (x + y) (x + y) (x + y) (x + y)
Hence, f (x, y) = (x + y) (x + y)
[f (x, y)] = [(x + y) (x + y)]
= xy + xy
which is the required DNF.
SECTION – B
2. Prove that n3 + 2n is divisible by 3 using principle of
mathematical induction, where n is natural number.
Ans. Let S(n) : n3 + 2n is divisible by 3.
Step I : Inductive base : For n = 1
(1)3 + 2.1 = 3 which is divisible by 3
Thus, S(1) is true.
Step II : Inductive hypothesis : Let S(k) is true i.e., k3 + 2k is
divisible by 3 holds true.
or k3 + 2k = 3s for s N
SP–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Step III : Inductive step : We have to show that S(k + 1) is true
i.e., (k + 1)3 + 2(k + 1) is divisible by 3
Consider (k + 1)3 + 2(k + 1)
= k3 + 1 + 3k2 + 3k + 2k + 2
= (k3 + 2k) + 3(k2 + k + 1)
= 3s + 3l where l = k2 + k + 1 N
= 3(s + l)
Therefore, S(k + 1) is true
Hence by principle of mathematical induction S(n) is true for all
n N.
3. Solve the recurrence relation using generating function :
an – 7an – 1 + 10an – 2 = 0 with a0 = 3, a1 = 3.
Ans. an – 7an – 1 + 10 an – 2 = 0,
Let in assume n 2
Multiply by xn and take sum from 2 to .
n 2
n2
n 2
an xn 7 an –1 xn 10 an2 xn 0
(a2 x2 + a3 x3 + a4 x4 + ....) – 7 (a1 x2 + a2 x3 + ....)
+ 10 (a0 x2 + a1 x3 + ....) = 0
We know that
G(x) =
an xn a0 a1 x ....
n0
G(x) – a0 – a1 x – 7 x (G(x) – a0) + 10 x2 G(x) = 0
G(x) [1 – 7 x + 10 x2] = a0 + a1 x – 7 a0 x
= 3 + 3x – 21 x = 3 – 18 x
3 18 x
3 18 x
10 x 2 7 x 1 10 x 2 5 x 2 x 1
3 18 x
3 18 x
=
5 x (2 x 1) 1 (2 x 1) (5 x – 1)(2 x 1)
G(x) =
Now,
3 18 x
A
B
=
(5 x 1) (2 x 1) 5 x 1 2 x 1
3 – 18 x = A (2 x – 1) + B (5 x – 1)
1
put
x=
2
5
3
3 – 9 = B 1 – 6 =
B B = – 4
2
2
1
put
x=
5
3
3
2
18
3–
= A 1 A 1 A = 1
5
5
5
5
SP–8 C (CS/IT-Sem-3)
Solved Paper (2015-16)
1
4
4
1
=
5 x 1 2x 1
1 2x 1 5x
an = 4.2n – 5n
G(x) =
4. Express the following statements using quantifiers and
logical connectives.
a. Mathematics book that is published in India has a blue
cover.
b. All animals are mortal. All human being are animal.
Therefore, all human being are mortal.
c. There exists a mathematics book with a cover that is not
blue.
d. He eats crackers only if he drinks milk.
e. There are mathematics books that are published outside
India.
f. Not all books have bibliographies.
Ans.
a. P(x) : x is a mathematic book published in India
Q(x) : x is a mathematic book of blue cover
x P(x) Q(x).
b. P(x) : x is an animal
Q(x) : x is mortal
x P(x) Q(x)
R(x) : x is a human being
x R(x) P(x).
c. P(x) : x is a mathematics book
Q(x) : x is not a blue color
x, P(x) Q(x).
d. P(x) : x drinks milk
Q(x) : x eats crackers
for x, if P(x) then Q(x).
or x, P(x) Q(x).
e. P(x) : x is a mathematics book
Q(x) : x is published outside India
x P(x) Q(x).
f. P(x) : x is a book having bibliography ~ x, P(x).
5. Draw the Hasse diagram of [P (a, b, c), ] (Note : ‘’ stands
for subset). Find greatest element, least element, minimal
element and maximal element.
Ans. Let a1 and a2 be two complements of an element a L.
Then by definition of complement
a a1 I
a a1 0
...(1)
Discrete Structures & Theory of Logic
SP–9 C (CS/IT-Sem-3)
a a2 I
...(2)
a a2 0
Consider
a1 = a1 0
= a1 (a a2)
[from (2)]
= (a1 a) (a1 a2)
[Distributive property]
= (a a1) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (1)]
= a1 a2
...(3)
Now Consider a2 = a2 0
= a2 (a a1)
[from (2)]
= (a2 a) (a2 a1)
[Distributive property]
= (a a2) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (1)]
= a1 a2
...(4)
Hence, from (3) and (4),
a1 = a2
So, for bounded distributive lattice complement is unique.
Hasse diagram of [P(a, b, c), ] is shown in Fig. 1.
{a, b, c}
{a, b}
{a, c}
{a}
{b}
{b, c}
{c}
Fig. 1.
Greatest element is {a, b, c} and maximal element is {a, b, c}.
The least element is and minimal element is .
6. Simplify the following boolean expressions using k-map :
a. Y = ((AB) + A + AB)
b. A B C D + A B C D + A B CD + A B CD = A B
Ans.
a.
Y = ((AB) + A + AB)
= ((AB)) (A + (AB))
= (AB) ((A) (AB))
SP–10 C (CS/IT-Sem-3)
Solved Paper (2015-16)
= AB (A (A + B))
= AB (AA + AB)
= AB(0 + AB) = AB AB
= ABB
=0
Here, we find that the expression is not in minterm. For getting
minterm, we simplify and find that its value is already zero. Hence,
no need to use K-map for further simplification.
b. A B C D + A B C D + A B CD + A B CD = A B
= A BCD + ABCD + ABCD + ABCD
CD AB
ABCD
1
ABCD
1
ABCD
1
ABCD
1
Fig. 2.
On simplification by K-map, we get AB corresponding to all the
four one’s.
7. Let G be the set of all non-zero real number and let
a*b = ab/2. Show that (G*) be an abelian group.
Ans.
i. Closure property : Let a, b G.
a*b=
ab
G as ab 0
2
* is closure in G.
ii. Associativity : Let a, b, c G
bc a(bc) abc
Considera * (b * c) = a *
2
4
4
ab
(ab) c abc
(a * b) * c = * c =
2
4
4
* is associative in G.
iii. Existence of the identity : Let a G and e such that
ae
=a
2
ae = 2a
e=2
2 is the identity element in G.
iv. Existence of the inverse : Let a G and b G such that a * b =
e=2
ab
=2
2
a*e=
Discrete Structures & Theory of Logic
SP–11 C (CS/IT-Sem-3)
ab = 4
4
b=
a
4
, a G.
a
v. Commutative : Let a, b G
The inverse of a is
ab
2
ba ab
and
b*a=
2
2
* is commutative.
Thus, (G, *) is an abelian group.
a*b=
8. The following relation on A = {1, 2, 3, 4}. Determine whether
the following :
a. R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}.
b. R = A × A
Is an equivalence relation or not.
Ans.
a. R = {(1, 3), (3, 1), (1, 1), (1, 2), (3, 3), (4, 4)}
Reflexive : (a, a) R a A
(1, 1) R, (2, 2) R
R is not reflexive.
Symmetric : Let (a, b) R then (b, a) R.
(1, 3) R so (3, 1) R
(1, 2) R but (2, 1) R
R is not symmetric.
Transitive : Let (a, b) R and (b, c) R then (a, c) R
(1, 3) R and (3, 1) R so (1, 1) R
(2, 1) R and (1, 3) R but (2, 3) R
R is not transitive.
Since, R is not reflexive, not symmetric, and not transitive so R is
not an equivalence relation.
b. R = A × A
Since, A × A contains all possible elements of set A. So, R is reflexive,
symmetric and transitive. Hence R is an equivalence relation.
9. If the permutation of the elements of {1, 2, 3, 4, 5} are given
by a = (1 2 3) (4 5), b = (1) (2) (3) (4 5), c = (1 5 2 4) (3). Find the
value of x, if ax = b. And also prove that the set Z4 = (0, 1, 2, 3)
is a commutative ring with respect to the binary modulo
operation +4 and *4.
Ans. ax = b (123) (45) x = (1) (2) (3) (4, 5)
1 2 3 4 5 1 2 3 4 5
1 2 3 4 5
x
2 3 1 4 5 1 2 3 5 4
1 2 3 5 4
SP–12 C (CS/IT-Sem-3)
Solved Paper (2015-16)
1 2 3 4 5
1 2 3 4 5
2 3 1 5 4 x = 1 2 3 5 4
1 2 3 4 5
x=
2 3 1 5 4
1
1 2 3 4 5
1 2 3 5 4
2 3 1 5 4 1 2 3 4 5
=
1 2 3 4 5 1 2 3 5 4
1 2 3 4 5 1 2 3 4 5
=
3 1 2 5 4 1 2 3 5 4
x=
4
0
1
i.
ii.
iii.
iv.
v.
vi.
0 1 2 3
0 1 2 3
1 2 3 0
1 2 3 4 5
3 1 2 4 5
4
0
1
2
3
0
0
0
0
0
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
2 2 3 0 1
3 3 0 1 2
We find from these tables :
All the entries in both the tables belong to Z4. Hence, Z4 is closed
with respect to both operations.
Commutative law : The entries of 1st, 2nd, 3rd, 4th rows are identical
with the corresponding elements of the 1st, 2nd, 3rd, 4th columns
respectively in both the tables. Hence, Z4 is commutative with respect
to both operations.
Associative law : The associative law for addition and multiplication
a +4 (b +4 c) = (a +4 b) +4 c for all a, b, c Z4
a ×4 (b ×4 c) = (a ×4 b) ×4 c, for all a, b, c Z4
can easily be verified.
Existence of identity : 0 is the additive identity and 1 is
multiplicative identity for Z4.
Existence of inverse : The additive inverses of 0, 1, 2, 3 are 0, 3, 2,
1 respectively.
Multiplicative inverse of non-zero element 1, 2, 3 are 1, 2, 3
respectively.
Distributive law : Multiplication is distributive over addition i.e.,
a ×4(b +4c) = a ×4b + a ×4c
(b +4c) ×4a = b ×4a + c ×4a
For, a ×4(b +4c) = a ×4(b + c) for b +4 c = b + c (mod 4)
= least positive remainder when a × (b + c) is
divided by 4
Discrete Structures & Theory of Logic
SP–13 C (CS/IT-Sem-3)
= least positive remainder when ab + ac is divided by 4
= ab +4ac
= a ×4b +4a ×4c
For a ×4b = a × b (mod 4)
Since (Z4, +4) is an abelian group, (Z4, ×4) is a semigroup and the
operation is distributive over addition. The (Z4, +4, ×4) is a ring. Now
(Z4, ×4 ) is commutative with respect to ×4 . Therefore, it is a
commutative ring.
SECTION – C
10. Let L be a bounded distributed lattice, prove if a complement
exists, it is unique. Is D12 a complemented lattice ? Draw
the Hasse diagram of [P (a, b, c), <], (Note : ‘<’ stands for
subset). Find greatest element, least element, minimal
element and maximal element.
Ans. Let a1 and a2 be two complements of an element a L.
Then by definition of complement
a a1 I
a a1 0
...(1)
a a2 I
...(2)
a a2 0
Consider
a1 = a1 0
= a1 (a a2)
[from (2)]
= (a1 a) (a1 a2)
[Distributive property]
= (a a1) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (1)]
= a1 a2
...(3)
Now Consider a2 = a2 0
= a2 (a a1)
[from (2)]
= (a2 a) (a2 a1)
[Distributive property]
= (a a2) (a1 a2)
[Commutative property]
= I (a1 a2)
[from (1)]
= a1 a2
...(4)
Hence, from (3) and (4),
a1 = a2
So, for bounded distributive lattice complement is unique.
Hasse diagram of [P(a, b, c), ] is shown in Fig. 3.
SP–14 C (CS/IT-Sem-3)
Solved Paper (2015-16)
{a, b, c}
{a, b}
{a, c}
{a}
{b}
{b, c}
{c}
Fig. 3.
Greatest element is {a, b, c} and maximal element is {a, b, c}.
The least element is and minimal element is .
11. Determine whether each of these functions is a bijective
from R to R.
a. f(x) = x2 + 1
b. f(x) = x3
c. f(x) = (x2 + 1)/(x2 + 2)
Ans.
a. f(x) = x2 + 1
Let x1, x2 R such that
f(x1) = f(x2 )
x 12 + 1 = x 22 + 1
x12 = x22
x1 = ± x2
Therefore, if x2 = 1 then x1 = ± 1
So, f is not one-to-one.
Hence, f is not bijective.
b. Let x1, x2 R such that f(x1) = f(x2)
x13 = x23
x1 = x2
f is one-to-one.
Let y R such that
y = x3
x = (y)1/3
For y R a unique x R such that y = f (x)
f is onto.
Hence, f is bijective.
c. Let x1, x2 R such that f(x1) = f(x2)
x12 1
x2 1
= 22
2
x1 2
x2 2
SP–15 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
If
x1 = 1, x2 = – 1 then f(x1) = f(x2)
but
x1 x2
f is not one-to-one.
Hence, f is not bijective.
12. a. Prove that inverse of each element in a group is unique.
b. Show that G = [(1, 2, 4, 5, 7, 8), ×9] is cyclic. How many
generators are there ? What are they ?
Ans.
a. Let (if possible) b and c be two inverses of element a G.
Then by definition of group :
b*a= a*b=e
and
a*c= c*a=e
where e is the identity element of G
Now
b = e * b = (c * a) * b
= c * (a * b)
= c*c
=c
b=c
Therefore, inverse of an element is unique in (G, *).
b. Composition table for X9 is
X9
1
2
4
5
7
8
1
2
4
5
7
8
2
2
4
5
7
8
3
4
8
1
5
7
4
8
7
2
1
5
5
1
2
7
8
4
7
5
1
8
4
2
8
7
5
4
2
1
1 is identity element of group G
21 = 2 2 mod 9
22 = 4 4 mod 9
23 = 8 8 mod 9
24 = 16 7 mod 9
25 = 32 5 mod 9
26 = 64 1 mod 9
Therefore, 2 is generator of G. Hence G is cyclic.
Similarly, 5 is also generator of G.
Hence there are two generators 2 and 5.
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B.Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2016-17
DISCRETE STRUCTURES AND
GRAPH THEORY
Time : 3 Hours
Max. Marks : 100
Section-A
Note : Attempt all parts. All parts carry equal marks. Write answer of
each part in short.
(2 × 10 = 20)
1. a. Let R be a relation on the set of natural numbers N, as
R = {(x, y) : x, y N, 3x + y = 19}. Find the domain and range of
R. Verify whether R is reflexive.
b. Show that the relation R on the set Z of integers given by
R = {(a, b) : 3 divides a – b}, is an equivalence relation.
c. Show the implications without constructing the truth table
(P Q) Q P Q.
d. Show that the “greater than or equal” relation (>=) is a
partial ordering on the set of integers.
e. Distinguish between bounded lattice and complemented
lattice.
f. Find the recurrence relation from yn = A2n + B(–3)n.
g. Define ring and give an example of a ring with zero divisors.
h. State the applications of binary search tree.
i. Define multigraph. Explain with example in brief.
j. Let G be a graph with ten vertices. If four vertices has degree
four and six vertices has degree five, then find the number
of edges of G.
SP–2 C (CS/IT-Sem-3)
Solved Paper (2016-17)
Section-B
Note : Attempt any five questions from this section.
(10 × 5 = 50)
2. Write the s ymbolic form and negate the following
statements :
• Everyone who is healthy can do all kinds of work.
• Some people are not admired by everyone.
• Everyone should help his neighbours, or his neighbours
will not help him.
3. In a lattice if a b c, then show that
a. a b = b c
b. (a b) (b c) = (a b) (a c) = b
4. State and prove Lagrange’s theorem for group. Is the
converse true ?
5. Prove that a simple graph with n vertices and k components
can have at most
(n k) ( n k 1)
edges.
2
6. Prove by induction :
n
1
1
1
=
.
+
+ ... +
1.2 2.3
n ( n + 1) (n + 1)
7. Solve the recurrence relation yn + 2 – 5yn + 1 + 6yn = 5n subject
to the condition y0 = 0, y1 = 2.
8. a. Prove that every finite subset of a lattice has an LUB and a
GLB.
b. Give an example of a lattice which is a modular but not a
distributive.
9. Explain in detail about the binary tree traversal with an
example.
Section-C
Note : Attempt any two questions from this section.
(15 × 2 = 30)
10. a. Prove that a connected graph G is Euler graph if and only if
every vertex of G is of even degree.
b. Which of the following simple graph have a Hamiltonian
circuit or, if not a Hamiltonian path ?
SP–3 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
a
b
a
b
e
c
d
c
d
G1
G2
a
b
d
c
g
e
f
G3
Fig. 1.
11. a. Find the Boolean algebra expression for the following
system.
A
B
C
Fig. 2.
b. Suppose that a cookie shop has four different kinds of
cookies. How many different way can six cookies be chosen ?
12. a. Prove that every cyclic group is an abelian group.
b. Obtain all distinct left cosets of {(0), (3)} in the group (Z6, +6)
and find their union.
c. Find the left cosets of {[0], [3]} in the group (Z6, +6).
SP–4 C (CS/IT-Sem-3)
Solved Paper (2016-17)
SOLUTION OF PAPER (2016-17)
Section-A
Note : Attempt all parts. All parts carry equal marks. Write answer of
each part in short.
(2 × 10 = 20)
1. a. Let R be a relation on the set of natural numbers N, as
R = {(x, y) : x, y N, 3x + y = 19}. Find the domain and range of
R. Verify whether R is reflexive.
Ans. By definition of relation,
R = {(1, 16), (2, 13), (3, 10), (4, 7), (5, 4), (6, 1)}
Domain = {1, 2, 3, 4, 5, 6}
Range = {16, 13, 10, 7, 4, 1}
R is not reflexive since (1, 1) R.
b. Show that the relation R on the set Z of integers given by
R = {(a, b) : 3 divides a – b}, is an equivalence relation.
Ans. Reflexive : a – a = 0 is divisible by 3
(a, a) R a Z
R is reflexive.
Symmetric : Let (a, b) R a – b is divisible by 3
– (a – b) is divisible by 3
b – a is divisible by 3
(b, a) R
R is symmetric.
Transitive : Let (a, b) R and (b, c) R
a – b is divisible by 3 and b – c is divisible by 3
Then a – b + b – c is divisible by 3
a – c is divisible by 3
(a, c) R
R is transitive.
Hence, R is equivalence relation.
c. Show the implications without constructing the truth table
(P Q) Q P Q.
Ans. (P Q) Q P Q
Take L.H.S
(P Q) Q = (~P Q) Q
= (~(~P Q)) Q
= (P ~ Q) Q
= (P Q) (~ Q Q)
= (P Q) T = P Q
It is equivalent.
SP–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
d. Show that the “greater than or equal” relation (>=) is a
partial ordering on the set of integers.
Ans. Reflexive :
a a a Z (set of integer)
(a, a) A
R is reflexive.
Antisymmetric : Let (a, b) R and (b, a) R
a b and b a
a=b
R is antisymmetric.
Transitive : Let (a, b) R and (b, c) R
a b and b c
a c (a, c) R
R is transitive.
Hence, R is partial order relation.
e. Distinguish between bounded lattice and complemented
lattice.
Ans. Bounded lattice : A lattice which has both elements 0 and 1 is
called a bounded lattice.
Complemented lattice : A lattice L is called complemented lattice
if it is bounded and if every element in L has complement.
f. Find the recurrence relation from yn = A2n + B(–3)n.
yn = A2n + B (– 3)n
Ans. Given :
Therefore, yn + 1 = A(2)n + 1 + B (– 3)n + 1
= 2A (2)n – 3B (– 3)n
and
yn + 2 = A(2)n + 2 + B (– 3)n + 2
= 4A (2)n + 9B (– 3)n
Eliminating A and B from these equations, we get
yn
1
1
yn1 2 – 3 = 0
yn 2 4 9
= yn + 2 – yn + 1 – 6yn = 0 which is the
required recurrence relation.
g. Define ring and give an example of a ring with zero divisors.
Ans. Ring : A non-empty set R is a ring if it is equipped with two binary
operations called addition and multiplication and denoted by '+'
and '.' respectively i.e., for all a, b R we have a + b R and a.b R
and it satisfies the following properties :
i. Addition is associative i.e.,
(a + b) + c = a + (b + c) a, b, c R
ii. Addition is commutative i.e.,
a + b = b + a a, b R
SP–6 C (CS/IT-Sem-3)
Solved Paper (2016-17)
iii. There exists an element 0 R such that
0 + a = a = a + 0, a R
iv. To each element a in R there exists an element – a in R such that
a + (– a) = 0
v. Multiplication is associative i.e.,
a.(b.c) = (a.b).c, a b, c R
vi. Multiplication is distributive with respect to addition i.e., for all
a, b, c R,
Example of ring with zero divisors : R = {a set of 2 × 2
matrices}.
Field : A ring R with at least two elements is called a field if it has
following properties :
i. R is commutative
ii. R has unity
iii. R is such that each non-zero element possesses multiplicative
inverse.
For example : The rings of real numbers and complex numbers
are also fields.
h. State the applications of binary search tree.
Ans. One of the most common applications is to efficiently store data in
sorted form in order to access and search stored elements quickly.
For example, std :: map or std :: set in C++ Standard Library.
Binary tree as data structure is useful for various implementations
of expression parsers and expression solvers.
i. Define multigraph. Explain with example in brief.
Ans. A multigraphs G(V, E) consists of a set of vertices V and a set of
edges E such that edge set E may contain multiple edges and self
loops.
Example :
a. Undirected multigraph :
v1
e1
v4
e2
v2
e6
e5
e3
e4
Fig. 1.
v3
SP–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
b. Directed multigraph :
e3
v1
e1
v4
e2
e7
v2
e5
e4
e6
Fig. 2.
v3
j. Let G be a graph with ten vertices. If four vertices has degree
four and six vertices has degree five, then find the number
of edges of G.
Ans. We know that
deg (vi ) = 2e
i
4 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 5 = 2e
16 + 30 = 2e
2e = 46
e = 23
Section-B
Note : Attempt any five questions from this section.
(10 × 5 = 50)
2. Write the s ymbolic form and negate the following
statements :
• Everyone who is healthy can do all kinds of work.
• Some people are not admired by everyone.
• Everyone should help his neighbours, or his neighbours
will not help him.
Ans.
a. Symbolic form :
Let P(x): x is healthy and Q(x): x do all work
x(P(x) Q(x))
Negation : ( x (P(x) Q(x))
b. Symbolic form :
Let P(x) : x is a person
A(x, y) : x admires y
The given statement can be written as “There is a person who is not
admired by some person” and it is (x) ( y)[P(x) P(y) A(x, y)]
Negation : ( x) ( y) [P (x) P(y) A(x, y)]
c. Symbolic form :
Let N(x, y) : x and y are neighbours
H(x, y) : x should help y
P(x, y) : x will help y
The statement can be written as “For every person x and every
person y, if x and y are neighbours, then either x should help y or y
SP–8 C (CS/IT-Sem-3)
Solved Paper (2016-17)
will not help x” and it is ( x) ( y)[N(x, y) (H(x, y) P(y, x))]
Negation : ( x) ( y) [N(x, y) (H (x, y) P (y, x))]
3. In a lattice if a b c, then show that
a. a b = b c
b. (a b) (b c) = (a b) (a c) = b
Ans.
a. Given : a b c
Now
a b = least upper bound of a, b
= least {all upper bounds of a, b}
= least {b, c, ...}
[using a b c]
=b
...(1)
and
b c = greatest lower bound of b, c
= maximum {all lower bounds of b, c}
= maximum {b, a, ...}
[using a b c]
=b
...(2)
Eq. (1) and (2) gives, a b = b c
b. (a b) (b c) (a b) (a c) = b
Consider, (a b) (b c)
= b b [using a b c and definition of and ]
=b
...(3)
and(a b) (a c) =
bc
=b
...(4)
From eq. (3) and (4), (a b) (b c) = (a b) (a c) = b.
4. State and prove Lagrange’s theorem for group. Is the
converse true ?
Ans. Lagrange’s theorem :
Statement : The order of each subgroup of a finite group is a
divisor of the order of the group.
Proof : Let G be a group of finite order n. Let H be a subgroup of G
and let O(H) = m. Suppose h1, h2.... ..., hm are the m members of H.
Let a G, then Ha is the right coset of H in G and we have
Ha = {h1 a, h2 a,....hm a}
Ha has m distinct members, since = hi a = hj a hi = hj
Therefore, each right coset of H in G has m distinct members. Any
two distinct right cosets of H in G are disjoint i.e., they have no
element in common. Since G is a finite group, the number of distinct
right cosets of H in G will be finite, say, equal to k. The union of
these k distinct right cosets of H in G is equal to G.
Thus, if Ha1, Ha2,...., Hak are the k distinct right cosets of H in G.
Then G = Ha1 Ha2 Ha3 .... Hak
the number of elements in G = the number of elements in
Ha1 + ...... + the number of elements in Ha2 +......+ the number of
elements in Hak
Discrete Structures & Theory of Logic
SP–9 C (CS/IT-Sem-3)
O(G) = km
n = km
n
k=
m
m is a divisor of n.
O(H) is a divisor of O(G).
Proof of converse : If G be a finite group of order n and n G,
then
an = e
Let o (a) = m which implies am = e.
Now, the subset H of G consisting of all the integral power of a is a
subgroup of G and the order of H is m.
Then, by the Lagrange’s theorem, m is divisor of n.
Let n = mk, then
an = amk = (am)k = ek = e
Yes, the converse is true.
5. Prove that a simple graph with n vertices and k components
(n k) ( n k 1)
edges.
2
Ans. Let the number of vertices in each of the k-components of a graph
G be n1, n2, ...., nk, then we get
n1 + n2 + ... + nk = n where ni 1 (i = 1, 2, ..., k)
can have at most
k
k
Now,
(ni 1) =
i 1
i
i 1
i 1
2
k
k
n 1 = n – k
2
2
(ni 1) = n + k – 2nk
i 1
k
or
k
(n
i
i 1
1)2 2 (ni 1)(n j 1) = n2 + k2 – 2nk
i 1 j 1
i j
k
or
(n
i
1)2 + 2(non-negative terms) = n2 + k2 – 2nk
i 1
[ ni – 1 0, nj – 1 0]
k
or
(n
i
2
1) n2 + k2 – 2nk
i 1
k
or
k
k
2
i
n 1 2 n
i 1
i
i 1
n2 + k2 – 2nk
i 1
k
or
2
i
k 2n n2 + k2 – 2nk
2
i
n n2 + k2 – 2nk – k + n
n
i 1
k
or
n
i 1
SP–10 C (CS/IT-Sem-3)
Solved Paper (2016-17)
= n(n – k + 1) – k (n – k + 1)
= (n – k)(n – k + 1)
...(1)
We know that the maximum number of edges in the ith component
of
ni (ni 1)
2
Therefore, the maximum number of edges in G is :
G=
ni
C2 =
1
1
1
ni (ni 1) = 2 ni2 ni = 2 ni2 n
2
1
(n – k)(n – k + 1) by using eq. (1)
2
6. Prove by induction :
n
1
1
1
=
.
+
+ ... +
1.2 2.3
n ( n + 1) (n + 1)
Ans. Let the given statement be denoted by S(n).
1. Inductive base : For n = 1
1
1
1
=
1
1
2
1.2
Hence S (1) is true.
2. Inductive hypothesis : Assume that S (k) is true i.e.,
1
1
1
k
.....
=
1.2 2.3
k(k 1) k 1
3. Inductive step : We wish to show that the statement is true for
n = k + 1 i.e.,
1
1
1
k1
.....
=
1.2 2.3
(k 1)(k 2)
k2
Now,
1
1
1
1
.....
1.2 2.3
k(k 1) ( k 1)(k 2)
=
k
1
k2 2k 1
k 1 ( k 1)(k 2) ( k 1)(k 2)
k1
k2
Thus, S(k + 1) is true whenever S(k) is true. By principle of
mathematical induction, S(n) is true for all positive integer n.
=
7. Solve the recurrence relation yn + 2 – 5yn + 1 + 6yn = 5n subject
to the condition y0 = 0, y1 = 2.
Ans.
Let G(t) =
a t
n
n 0
n
be generating function of sequence {an}.
SP–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Multiplying given equation by tn and summing from
n = 0 to , we have
an 2 t n – 5
n 0
n 0
G(t) – a0 – a1 t
t2
Put
an 1 t n 6
n 0
5
n
tn
n 0
1
G (t) – a0
– 5
6 G(t) 1 – 5t
t
a0 = 0 and a1 = 2
G(t) – 2t – 5t G(t) + 6t2 G(t) =
G(t) – 5t G(t) + 6t2 G(t) =
t2
1 – 5t
t2
+ 2t
1 – 5t
G(t) (1 – 5t + 6t2) =
t2
+ 2t
1 – 5t
(6t2 – 5t + 1) G(t) =
t2
+ 2t
1 – 5t
G(t) =
an t n
t2
2t
(1 – 5t)(3t – 1)(2t – 1) (3t – 1)(2t – 1)
t2
2t
(1 – 5t)(1 – 3t)(1 – 2t) (1 – 3t)(1 – 2t)
Let
=
A
B
C
t2
=
(1 – 5t) (1 – 3t) (1 – 2t)
(1 – 5t)(1 – 3t)(1 – 2t)
A = (1 – 5t)
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/5
=
t2
(1 – 3t)(1 – 2t) t 1/5
=
1 / 25
1
=
(1 – 3 / 5)(1 – 2 / 5)
6
B = (1 – 3t)
=
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/3
1/9
t2
=
3
–
5
3 – 2
(1 – 5t)(1 – 2t) t 1/3
3 3
= –
1
2
SP–12 C (CS/IT-Sem-3)
Solved Paper (2016-17)
C = (1 – 2t)
t2
(1 – 5t)(1 – 3t)(1 – 2t) t 1/2
t2
1/ 4
=
(2 – 5) (2 – 3)
(1 – 5t)(1 – 3t) t 1/2
2
2
1
=
3
=
Again,
2t
D
E
=
(1 – 3t)(1 – 2t)
(1 – 3t) (1 – 2t)
2t
D = (1 – 3t)
(1 – 3t)(1 – 2t) t 1/3
2t
2/3
2
=
(3 – 2)
(1 – 2t) t 1/3
3
2t
E = (1 – 2t)
(1 – 3t)(1 – 2t) t 1/ 2
=
2t
2/2
–2
=
2–3
(1 – 3t) t 1/ 2
2
1/ 6
1/ 2
1/3
2
2
G(t) =
–
–
(1 – 5t) (1 – 3t) (1 – 2t) (1 – 3t) (1 – 2t)
=
=
an t n =
n 0
an =
1/6
3/2
5/3
–
1 – 5t (1 – 3t) 1 – 2t
1
6
n 0
(5t)n
3
2
(3t) n –
n 0
5
(2t)n
3 n 0
1 n 3 n 5 n
(5) (3) – (2)
6
2
3
8. a. Prove that every finite subset of a lattice has an LUB and a
GLB.
b. Give an example of a lattice which is a modular but not a
distributive.
Ans.
a.
1. The theorem is true if the subset has 1 element, the element being
its own glb and lub.
2. It is also true if the subset has 2 elements.
Discrete Structures & Theory of Logic
SP–13 C (CS/IT-Sem-3)
3. Suppose the theorem holds for all subsets containing 1, 2, ..., k
elements, so that a subset a1, a2, ..., ak of L has a glb and a lub.
4. If L contains more than k elements, consider the subset
{a1, a2, ..., ak + 1} of L.
5. Let w = lub (a1, a2, ..., ak).
6. Let t = lub (w, ak + 1).
7. If s is any upper bound of a1, a2, ..., ak + 1, then s is each of a1, a2, ...,
ak and therefore s w.
8. Also, s ak + 1 and therefore s is an upper bound of w and ak + 1.
9. Hence s t.
10. That is, since t each a1, t is the lub of a1, a2, ..., ak + 1.
11. The theorem follows for the lub by finite induction.
12. If L is finite and contains m elements, the induction process stops
when k + 1 = m.
b.
1. The diamond is modular, but not distributive.
2. Obviously the pentagon cannot be embedded in it.
3. The diamond is not distributive :
y (x z) = y (y x) (y z) = 1
4. The distributive lattices are closed under sublattices and every
sublattice of a distributive lattice is itself a distributive lattice.
5. If the diamond can be embedded in a lattice, then that lattice has a
non-distributive sublattice, hence it is not distributive.
9. Explain in detail about the binary tree traversal with an
example.
Ans. Tree traversal : A traversal of tree is a process in which each
vertex is visited exactly once in a certain manner. For a binary tree
we have three types of traversal :
1. Preorder traversal : Each vertex is visited in the following order :
a. Visit the root (N).
b. Visit the left child (or subtree) of root (L).
c. Visit the right child (or subtree) of root (R).
2. Postorder traversal :
a. Visit the left child (subtree) of root.
b. Visit the right child (subtree) of root.
c. Visit the root.
3. Inorder traversal :
a. Visit the left child (subtree) of root.
b. Visit the root.
c. Visit the right child (subtree) of root.
SP–14 C (CS/IT-Sem-3)
Solved Paper (2016-17)
A binary tree with 12 vertices :
A
C
B
D
G
E
H
F
I
J
L
K
Fig. 3.
Preorder (NLR) : A B D H I E J C F K L G
Inorder (LNR)
: HDIBJEAKFLCG
Postorder (LRN) : H I D J E B K L F G C A
Section-C
Note : Attempt any two questions from this section.
(15 × 2 = 30)
10. a. Prove that a connected graph G is Euler graph if and only if
every vertex of G is of even degree.
b. Which of the following simple graph have a Hamiltonian
circuit or, if not a Hamiltonian path ?
a
b
a
b
e
c
d
c
d
G1
G2
a
b
d
c
g
e
f
G3
Fig. 4.
Ans.
a.
1. First of all we shall prove that if a non-empty connected graph is
Eulerian then it has no vertices of odd degree.
2. Let G be Eulerian.
3. Then G has an Eulerian trail which begins and ends at u.
4. If we travel along the trail then each time we visit a vertex. We use
two edges, one in and one out.
5. This is also true for the start vertex because we also end there.
6. Since an Eulerian trail uses every edge once, the degree of each
vertex must be a multiple of two and hence there are no vertices of
odd degree.
7. Now we shall prove that if a non-empty connected graph has no
vertices of odd degree then it is Eulerian.
8. Let every vertex of G have even degree.
9. We will now use a proof by mathematical induction on |E(G)|, the
number of edges of G.
Discrete Structures & Theory of Logic
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
b.
SP–15 C (CS/IT-Sem-3)
Basis of induction :
Let |E(G)| = 0, then G is the graph K1, and G is Eulerian.
Inductive step :
Let P(n) be the statement that all connected graphs on n edges of
even degree are Eulerian.
Assume P(n) is true for all n < |E(G)|.
Since each vertex has degree at least two, G contains a cycle C.
Delete the edges of the cycle C from G.
The resulting graph, G say, may not be connected.
However, each of its components will be connected, and will have
fewer than |E(G)| edges.
Also, all vertices of each component will be of even degree, because
the removal of the cycle either leaves the degree of a vertex
unchanged, or reduces it by two.
By the induction assumption, each component of G is therefore
Eulerian.
To show that G has an Eulerian trail, we start the trail at a vertex,
u say, of the cycle C and traverse the cycle until we meet a vertex,
c1 say, of one of the components of G .
We then traverse that component’s Eulerian trail, finally returning
to the cycle C at the same vertex, c1.
We then continue along the cycle C, traversing each component of
G as it meets the cycle.
Eventually, this process traverses all the edges of G and arrives
back at u, thus producing an Eulerian trail for G.
Thus, G is Eulerian by the principle of mathematical induction.
G1 : The graph G1 shown in Fig. 4 contains Hamiltonian circuit, i.e.,
a – b – c – d – e – a and also a Hamiltonian path, i.e., abcde.
G2 : The graph G2 shown in Fig. 4 does not contain Hamiltonian
circuit since every cycle containing every vertex must contain the
edge e twice. But the graph does have a Hamiltonian path a – b – c
– d.
G3 : The graph G3 shown in Fig. 4 neither have Hamiltonian circuit
nor have Hamiltonian path because any traversal does not cover all
the vertices.
11. a. Find the Boolean algebra expression for the following
system.
A
B
C
Fig. 5.
SP–16 C (CS/IT-Sem-3)
Ans.
A
B
C
Solved Paper (2016-17)
ABC
ABC + A(B + C)
A.(B + C)
B
B + C
C
Fig. 6.
b. Suppose that a cookie shop has four different kinds of
cookies. How many different way can six cookies be chosen ?
Ans. As the order in which each cookie is chosen does not matter and
each kind of cookies can be chosen as many as 6 times, the number
of ways these cookies can be chosen is the number of 6-combination
with repetition allowed from a set with 4 distinct elements.
The number of ways to choose six cookies in the bakery shop is the
number of 6 combinations of a set with four elements.
C(4 + 6 – 1, 6) = C(9, 6)
Since
C(9, 6) = C(9, 3) = (9 · 8 · 7) / (1 · 2 · 3) = 84
Therefore, there are 84 different ways to choose the six cookies.
12. a. Prove that every cyclic group is an abelian group.
b. Obtain all distinct left cosets of {(0), (3)} in the group (Z6, +6)
and find their union.
c. Find the left cosets of {[0], [3]} in the group (Z6, +6).
Ans.
a. Let G be a cyclic group and let a be a generator of G so that
G = <a> = {an : n Z}
If g1 and g2 are any two elements of G, there exist integers r and s
such that g1 = ar and g2 = as. Then
g1 g2 = ar as = ar+s = as+r = as . ar = g2g1
So, G is abelian.
b. [0] + H = [3] + H, [1] + [4] + H and [2] + H = [5] + H are the three
distinct left cosets of H in (Z6 , + 6 ).
We would have the following left cosets :
g1H = {g1 h, h H}
g2H = {g2 h, h H}
gnH = {gn h, h H}
The union of all these sets will include all the g s, since for each set
gk = {gk h, h H}
we have ge gk = {gk h, h H}
where e is the identity.
Then if we make the union of all these sets we will have at least all
the elements of g. The other elements are merely gh for some h.
Discrete Structures & Theory of Logic
SP–17 C (CS/IT-Sem-3)
But since ghG they would be repeated elements in the union. So,
the union of all left cosets of H in G is G, i.e.,
Z6 = {[0], [1], [2], [3], [4], [5]}
c. Let
Z6 = {[0], [1], [2], [3], [4], [5]} be a group.
H = {[0], [3]} be a subgroup of (Z6, +6).
The left cosets of H are,
[0] + H = {[0], [3]}
[1] + H = {[1], [4]}
[2] + H = {[2], [5]}
[3] + H = {[3], [0]}
[4] + H = {[4], [1]}
[5] + H = {[5], [2]}
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B.Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2017-18
DISCRETE STRUCTURES &
THEORY OF LOGIC
Time : 3 Hours
Note : 1.
2.
Max. Marks : 70
Attempt all Sections. If require any missing data; then choose
suitably.
Any special paper specific instructions.
SECTION – A
1. Attempt all questions in brief.
a. Define Eulerian path, circuit and graph.
(2 × 7 = 14)
b. Let A = (2, 4, 5, 7, 8) = B, aRb if and only if a + b <= 12. Find
relation matrix.
c. Explain edge colouring and k-edge colouring.
d. Define chromatic number and isomorphic graph.
e. Define union and intersection of multiset and find for
A = [1, 1, 4, 2, 2, 3], B = [1, 2, 2, 6, 3, 3]
f. Find the contrapositive of “If he has courage, then he will
win”.
g. Define rings and write its properties.
SECTION-B
2. Attempt any three of the following :
a. Prove by mathematical induction
3 + 33 + 333 + ............ 3333 = (10n + 1 – 9n – 10)/27
(7 × 3 = 21)
b. Define the following with one example :
i. Bipartite graph
ii. Complete graph
iii. How many edges in K7 and K3, 6 iv. Planar graph
c. For any positive integer D36, then find whether (D36, ‘|’) is
lattice or not ?
SP–2 C (CS/IT-Sem-3)
Solved Paper (2017-18)
d. Let X = {1, 2, 3 ...... 7} and R = {(x, y)|(x – y) is divisible by 3}. Is
R equivalence relation. Draw the digraph of R.
e. Simplify the following Boolean function using K-map :
F(x, y, z) = (0, 2, 3, 7)
SECTION-C
3. Attempt any one part of the following :
(7 × 1 = 7)
a. Solve ar – 6ar – 1 + 8ar – 2 = r.4r, given a0 = 8, and a1 = 1.
b. Show that : (r ~ q, r S, S ~ q, p q) ~ p are
inconsistent.
4. Attempt any one part of the following :
(7 × 1 = 7)
a. Write the properties of group. Show that the set (1, 2, 3, 4, 5)
is not group under addition and multiplication modulo 6.
b. Prove by mathematical induction
n4 – 4n2 is divisible by 3 for all n > = 2.
5. Attempt any one part of the following :
(7 × 1 = 7)
a. Explain modular lattice, distribute lattice and bounded
lattice with example and diagram.
b. Draw the Hasse diagram of (A, ), where
A = {3, 4, 12, 24, 48, 72} and relation be such that a b if a
divides b.
6. Attempt any one part of the following :
(7 × 1 = 7)
a. Given the inorder and postorder traversal of a tree T :
Inorder : HFEABIGDC Postorder : BEHFACDGI
Determine the tree T and it's Preorder.
b. Translate the following sentences in quantified expressions
of predicate logic.
i. All students need financial aid. ii. Some cows are not white.
iii. Suresh will get if division if and only if he gets first div.
iv. If water is hot, then Shyam will swim in pool.
v. All integers are either even or odd integer.
7.
a.
1.
2.
3.
Attempt any one part of the following :
Define and explain any two the following :
BFS and DFS in trees
Euler graph
Adjacency matrix of a graph
(7 × 1 = 7)
b. Solve the recurrence relation : ar + 4ar – 2 + 4ar – 2 = r2.
Discrete Structures & Theory of Logic
SP–3 C (CS/IT-Sem-3)
SOLUTION OF PAPER (2017-18)
Note : 1.
2.
Attempt all Sections. If require any missing data; then choose
suitably.
Any special paper specific instructions.
SECTION – A
1. Attempt all questions in brief.
(2 × 7 = 14)
a. Define Eulerian path, circuit and graph.
Ans. Eulerian path : A path of graph G which includes each edge of G
exactly once is called Eulerian path.
Eulerian circuit : A circuit of graph G which include each edge of
G exactly once.
Eulerain graph : A graph containing an Eulerian circuit is called
Eulerian graph.
b. Let A = (2, 4, 5, 7, 8) = B, aRb if and only if a + b <= 12. Find
relation matrix.
Ans. R = {(2, 4), (2, 5), (2, 7), (2, 8) (4, 2), (4, 5), (4, 7), (4, 8) (5, 2), (5, 4),
(5, 7), (7, 2), (7, 4), (7, 5), (8, 2), (8, 4), (2, 2), (4, 4), (5, 5)}
2 4 5 7 8
2 1 1 1 1 1
4 1 1 1 1 1
mij = 5 1 1 1 1 0
7 1 1 1 0 0
8 1 1 0 0 0
c. Explain edge colouring and k-edge colouring.
Ans. Edge coloring : An edge coloring of a graph G may also be thought
of as equivalent to a vertex coloring of the line graph L(G), the
graph that has a vertex for every edge of G and an edge for every
pair of adjacent edges in G.
k-edge coloring : A proper edge coloring with k different colors is
called a (proper) k-edge coloring.
d. Define chromatic number and isomorphic graph.
Ans. Chromatic number : The minimum number of colours required
for the proper colouring of a graph so that no two adjacent vertices
have the same colour, is called chromatic number of a graph.
Isomorphic graph : If two graphs are isomorphic to each other
then :
i. Both have same number of vertices and edges.
SP–4 C (CS/IT-Sem-3)
Solved Paper (2017-18)
ii. Degree sequence of both graphs are same (degree sequence is the
sequence of degrees of the vertices of a graph arranged in nonincreasing order).
e. Define union and intersection of multiset and find for
A = [1, 1, 4, 2, 2, 3], B = [1, 2, 2, 6, 3, 3]
Ans. Union : Let A and B be two multisets. Then, A B, is the multiset
where the multiplicity of an element in the maximum of its
multiplicities in A and B.
Intersection : The intersection of A and B, A B, is the multiset
where the multiplicity of an element is the minimum of its
multiplicities in A and B.
Numerical :
A = {1, 1, 4, 2, 2, 3}, B = {1, 2, 2, 6, 3, 3}
Union :
A B = {1, 2, 3, 4, 6}
Intersection :
A B = {1, 2, 2, 3}
f. Find the contrapositive of “If he has courage, then he will
win”.
Ans. If he will not win then he does not have courage.
g. Define rings and write its properties.
Ans. Ring : A non-empty set R is a ring if it is equipped with two binary
i.
ii.
iii.
iv.
v.
vi.
operations called addition and multiplication and denoted by '+'
and '.' respectively i.e., for all a, b R we have a + b R and a.b R
and it satisfies the following properties :
Addition is associative i.e.,
(a + b) + c = a + (b + c) a, b, c R
Addition is commutative i.e.,
a + b = b + a a, b R
There exists an element 0 R such that
0 + a = a = a + 0, a R
To each element a in R there exists an element – a in R such that
a + (– a) = 0
Multiplication is associative i.e.,
a.(b.c) = (a.b).c, a b, c R
Multiplication is distributive with respect to addition i.e., for all
a, b, c R,
SECTION-B
2. Attempt any three of the following :
a. Prove by mathematical induction
3 + 33 + 333 + ............ 3333 = (10n + 1 – 9n – 10)/27
(7 × 3 = 21)
SP–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Ans. 3 + 33 + 333 + ............ + 3333 ... = (10n + 1 – 9n – 10)/27
Let given statement be denoted by S(n)
1. Inductive base : For n = 1
(102 9(1) 10)
100 19 81
3=
,3=
=3
27
27
27
3 = 3. Hence S(1) is tree.
2. Inductive hypothesis : Assume that S(k) is true i.e.,
3 + 33 + 333 + ............ + 3333 = (10k + 1 – 9k – 10)/27
3. Inductive steps : We have to show that S(k + 1) is also true i.e.,
3 + 33 + 333 + ............ (10k + 2 – 9(k + 1) – 10)/27
Now, 3 + 33 + ............ + 33 ............ 3
= 3 + 33 + 333 + ............ + 3 ............ 3
= (10k + 1 – 9k – 10)/27 + 3(10k + 1 – 1)/9
= (10k + 1 + 9k – 10 + 9.10k + 1 – 9)/27
= (10k + 1 + 9.10k + 1 – 9k – 8 – 10)/27 = (10k + 2 – 9(k + 1) – 10)/27
Thus S(k + 1) is true whenever S(k) is true. By the principle of
mathematical induction S(n) true for all positive integer n.
b. Define the following with one example :
i. Bipartite graph
ii. Complete graph
iii. How many edges in K7 and K3, 6 iv. Planar graph
Ans.
i. Bipartite graph : A graph G = (V, E) is bipartite if the vertex set
V can be partitioned into two subsets (disjoint) V1 and V2 such that
every edge in E connects a vertex in V1 and a vertex V2 (so that no
edge in G connects either two vertices in V1 or two vertices in V2).
(V1, V2) is called a bipartition of G.
x1
x2
x3
x4
y1
y2
x1
x2
x3
y1
y2
y3
x4
which
redrawn
as :
y3
Fig. 1. Some bipartite graphs.
ii. Complete graph : A simple graph, in which there is exactly one
edge between each pair of distinct vertices is called a complete
graph. The complete graph of n vertices is denoted by Kn. The
graphs K1 to K5 are shown below in Fig. 2.
K1
K2
Kn has exactly
K3
Fig. 2.
n(n – 1) n
= C2 edges
2
K4
K5
SP–6 C (CS/IT-Sem-3)
Solved Paper (2017-18)
iii. Number of edge in K7 : Since, Kn is complete graph with n vertices.
7(7 1) 7 6
Number of edge in K7 =
= 21
2
2
Number of edge in K3, 6 :
Since, Kn, m is a complete bipartite graph with n V1 and m V2
Number of edge in K3, 6 = 3 × 6 = 18
iv. Planar graph :
A graph G is said to be planar if there exists some geometric
representation of G which can be drawn on a plane such that no
two of its edges intersect except only at the common vertex.
i. A graph is said a planar graph, if it cannot be drawn on a plane
without a crossover between its edges crossing.
ii. The graphs shown in Fig. 3(a) and (b) are planar graphs.
(a)
Fig. 3. Some planar graph.
(b )
c. For any positive integer D36, then find whether (D36, ‘|’) is
lattice or not ?
Ans. D36 = Divisor of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hasse diagram :
(1 3) = {3, 6}, (1 2) = {2, 4}, (2 6) = {6, 18}, (9 4) = {}
36
12
12
6
4
9
3
2
1
Since,
9 4 = {}
So, D36 is not a lattice.
d. Let X = {1, 2, 3 ...... 7} and R = {(x, y)|(x – y) is divisible by 3}. Is
R equivalence relation. Draw the digraph of R.
X = {1, 2, 3, 4, 5, 6, 7}
Ans. Given that
and
R = {(x, y) : (x – y) is divisible by 3 }
Then R is an equivalence relation if
i. Reflexive : x X ( x x) is divisible by 3
So, ( x, x) X x X or, R is reflexive.
ii. Symmetric : Let x, y X and (x, y) R
(x – y) is divisible by 3 (x – y) = 3n1, (n1 being an integer)
SP–7 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
(y – x) = – 3n2 = 3n2, n2 is also an integer
So, y – x is divisible by 3 or R is symmetric.
iii. Transitive : Let x, y, z X and (x, y) R, (y, z) R
Then x – y = 3n1, y – z = 3n2, n1, n2 being integers
x – z = 3(n1 + n2), n1 + n2 = n3 be any integer
So, (x – z) is also divisible by 3 or (x, z) R
So, R is transitive.
Hence, R is an equivalence relation.
3
1
4
7
2
6
5
Fig. 4. Diagraph of R.
e. Simplify the following Boolean function using K-map :
F(x, y, z) = (0, 2, 3, 7)
Ans.
x
yz
0
00
01
1
1
11
10
1
1
1
F = x z yz
SECTION-C
3. Attempt any one part of the following :
(7 × 1 = 7)
a. Solve ar – 6ar – 1 + 8ar – 2 = r.4r, given a0 = 8, and a1 = 1.
Ans. ar – 6ar – 1 + 8ar – 2 = r4r
The characteristic equation is, x2 – 6x + 8 = 0, x2 – 2x – 4x + 8 = 0
(x – 2) (x – 4) = 0, x = 2, 4
The solution of the associated non-homogeneous recurrence relation is,
ar(h) = B1(2)r + B2(4)r
...(1)
Let particular solution of given equation is, ar(p) = r2(A0 + A1r)4r
Substituting in the given equation, we get
r2(A0 + A1r)4r – 6(r – 1)2 (A0 + A1(r – 1))4r – 1
+ 8(r – 2)2 (A0 + A1(r – 2)4r – 2 = r4r
6
r 2A 0 + A 1r 3 –
[(A0r2 – 2A0r + A0) + (A1r3 – A1 – 3A1r2 + 3A1r)2]
4
8
+ 2 [(A0r2 – 4rA0 + 4A0) + (A1r3 – 8A1 – 6A1r2 + 12A1r)] = r
4
SP–8 C (CS/IT-Sem-3)
Solved Paper (2017-18)
3
3
3
3
A r2 + 3A0r –
A –
A r3 +
A
2 0
2 0 2 1
2 1
9
9
1
+
A r2 –
A r+
A r2 – 2A0r + 2A0
2 1
2 1
2 0
1
A r3 – 4A1 – 3A1r2 – 6A1r = r
2 1
1
5
3
3
2A0r – A0r2 –
A –
A +
A r2 +
A r=r
2 0 2 1 2 1
2 1
Comparing both sides, we get
3
2A0 +
A =1
...(2)
2 1
A0 + 5A1 = 0
...(3)
2
10
Solving equation (2) and (3), we get A1 =
A0 =
17
17
To find the value of B1 and B2 put r = 0 and r = 1 in equation (1)
rA0 + A1r3 –
r = 0 a0 = B1 + B2
B1 + B2 = 8
r = 1 a1 = 2B1 + 4B2
...(4)
2B1 + 4B2 = 1
...(5)
31
Solving equations (4) and (5), we get B1 =
2
Complete solution is, ar = ar(h) + ar(p)
ar =
31 r 15 r
2
4 r2
2
2
15
B2 =
2
10 2 r
17 17 r 4
b. Show that : (r ~ q, r S, S ~ q, p q) ~ p are
inconsistent.
Ans. Following the indirect method, we introduce p as an additional premise
and show that this additional premise leads to a contradiction.
{1}
(1) p q
Rule P
{2}
(2) p
Rule P (assumed)
{1, 2}
(3) q
Rule T, (1), (2) and modus ponens
{4}
(4) s q
Rule P
{1, 2, 4}
(5) s
Rule T, (3), (4) and modus tollens
{6}
(6) r s
Rule P
{1, 2, 4, 6}
(7) r
Rule T, (5), (6) disjunctive syllogism
{8}
(8) r q
Rule P
{8}
(9) r q
Rule T, (8) and EQ16 (p q p q)
{8}
{1, 2, 4, 6}
(10) r q
(11) r q
Rule T, (8) and De Morgan’s law
Rule T, (7), (3) and conjunction
{1, 2, 4, 6, 8} (12) r q r q
Rule T, (10), (11) and conjunction.
Since, we know that set of formula is inconsistent if their conjunction
implies contradiction. Hence it leads to a contradiction. So, it is
inconsistent.
SP–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
4. Attempt any one part of the following :
(7 × 1 = 7)
a. Write the properties of group. Show that the set (1, 2, 3, 4, 5)
is not group under addition and multiplication modulo 6.
Ans. Properties of group :
Following are the properties of group :
1. a * b G a, b G [closure property]
2. a * (b * c) = (a * b) * c a, b, c G [associative property]
3. There exist an element e G such that for any a G
a * e = e * a = e [existence of identity]
4. For every a G, element a–1 G
such that a * a–1 = a–1 * a = e
For example : (Z, +), (R, +), and (Q, +) are all groups.
Numerical :
Addition modulo 6 (+6) : Composition table of S = {1, 2, 3, 4, 5}
under operation +6 is given as :
+6
1
2
3
4
5
1
2
3
4
5
0
2
3
4
5
0
1
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
Since, 1 +6 5 = 0 but 0 S i.e., S is not closed under addition modulo 6.
So, S is not a group.
Multiplication modulo 6 (*6) :
Composition table of S = {1, 2, 3, 4, 5} under operation *6 is given as
*6
1
2
3
4
5
1
1
2
3
4
5
2
2
4
0
2
4
3
3
0
3
0
3
4
4
2
0
4
2
5
5
4
3
2
1
Since, 2 *6 3 = 0 but 0 S i.e., S is not closed under multiplication
modulo 6.
So, S is not a group.
b. Prove by mathematical induction
n4 – 4n2 is divisible by 3 for all n > = 2.
Ans. Base case : If n = 0, then n4 – 4n2 = 0, which is divisible by 3.
Inductive hypothesis : For some n 0, n4 – 4n2 is divisible by 3.
Inductive step : Assume the inductive hypothesis is true for n.
SP–10 C (CS/IT-Sem-3)
Solved Paper (2017-18)
We need to show that (n + 1)4 – 4(n + 1)2 is divisible by 3. By the
inductive hypothesis, we know that n4 – 4n2 is divisible by 3.
Hence (n + 1)4 – 4(n + 1)2 is divisible by 3 if
(n + 1)4 – 4(n + 1)2 – (n4 –4n2) is divisible by 3.
Now (n + 1)4 – 4(n + 1)2 – (n4 – 4n2)
= n4 + 4n3 + 6n2 + 4n + 1 – 4n2 – 8n – 4 – n4 + 4n2
= 4n3 + 6n2 – 4n – 3,
which is divisible by 3 if 4n3 – 4n is. Since 4n3 – 4n = 4n(n + 1)
(n – 1), we see that 4n3 – 4n is always divisible by 3.
Going backwards, we conclude that (n + 1)4 – 4(n + 1)2 is divisible by
3, and that the inductive hypothesis holds for n + 1.
By the Principle of Mathematical Induction, n4 – 4n2 is divisible by 3,
for all n N.
5. Attempt any one part of the following :
(7 × 1 = 7)
a. Explain modular lattice, distribute lattice and bounded
lattice with example and diagram.
Ans. Modular distributive and bounded lattice :
Types of lattice :
1. Modular lattice : A lattice (L, ) is called modular lattice if,
a (b c) = (a b) c whenever a c for all a, b, c L.
2. Distributive lattice : A lattice L is said to be distributive if for
any element a, b and c of L following properties are satisfied :
i. a (b c) = (a b) (a c)
ii. a (b c) = (a b) (a c)
otherwise L is non-distributive lattice.
3. Bounded lattice : A lattice L is said to be bounded if it has a
greatest element 1 and a least element 0. In such lattice we have
a 1 = 1, a 1 = a
a 0 = a, a 0 = 0
a Land 0 a I
Example :
Let consider a Hasse diagram :
1
a
b
c
0
Fig. 5.
Modular lattice :
0 a i.e., taking b = 0
b (a c) = 0 0 = 0, a (b c) = a c = 0
Distributive lattice :
For a set S, the lattice P(S) is distributive, since union and intersection
each satisfy the distributive property.
SP–11 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Bounded lattice : Since, the given lattice has 1 as greatest and 0
as least element so it is bounded lattice.
b. Draw the Hasse diagram of (A, ), where
A = {3, 4, 12, 24, 48, 72} and relation be such that a b if a
divides b.
Ans. Hasse diagram of (A, ) where A = {3, 4, 12, 24, 48, 72}
1
a
c
b
0
Fig. 6.
6. Attempt any one part of the following :
(7 × 1 = 7)
a. Given the inorder and postorder traversal of a tree T :
Inorder : HFEABIGDC Postorder : BEHFACDGI
Determine the tree T and it's Preorder.
Ans. The root of tree is I.
I
Now elements on right of I are D, G, C and G comes last of all in
postorder traversal.
I
G
Now D and C are on right of G and D comes last of G and I postorder
traversal so
I
G
D
C
Now element left of I are H F E A B in inorder traversal and A
comes last of all in postorder traversal. Therefore tree will be
I
G
A
D
C
Now H F E are on left of A in inorder traversal and B comes last of all
and continuing in same manner. We will get final binary tree as
SP–12 C (CS/IT-Sem-3)
Solved Paper (2017-18)
I
G
A
H
D
B
F
C
E
Preorder traversal of above binary tree is CAFHEBDGI
b. Translate the following sentences in quantified expressions
of predicate logic.
i. All students need financial aid. ii. Some cows are not white.
iii. Suresh will get if division if and only if he gets first div.
iv. If water is hot, then Shyam will swim in pool.
v. All integers are either even or odd integer.
Ans.
i.
ii.
iii.
iv.
x [S(x) F(x)]
~ [(x) (C(x) W(x))]
Sentence is incorrect so cannot be translated into quantified expression.
W(x) : x is water
H(x) : x is hot
S(x) : x is Shyam
P(x) : x will swim in pool
x [((W(x) H(x)) (S(x) P(x))]
v. E(x) : x is even
O(x) : x is odd
x (E(x) O(x))
7.
a.
1.
2.
3.
Attempt any one part of the following :
Define and explain any two the following :
BFS and DFS in trees
Euler graph
Adjacency matrix of a graph
(7 × 1 = 7)
Ans.
1. Breadth First Search (BFS) : Breadth First Search (BFS) is an
algorithm for traversing or searching tree or graph data structures.
It starts at the tree root and explores the neighbour nodes first,
before moving to the next level neighbours.
Algorithmic steps :
Step 1 : Push the root node in the queue.
Step 2 : Loop until the queue is empty.
Step 3 : Remove the node from the queue.
Step 4 : If the removed node has unvisited child nodes, mark them
as visited and insert the unvisited children in the queue.
Depth First Search (DFS) :
Depth First Search (DFS) is an algorithm for traversing or searching
tree or graph data structures. One starts at the root (selecting
SP–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
some arbitrary node as the root in the case of a graph) and explores
as far as possible along each branch before backtracking.
Algorithmic steps :
Step 1 : Push the root node in the stack.
Step 2 : Loop until stack is empty.
Step 3 : Pick the node of the stack.
Step 4 : If the node has unvisited child nodes, get the unvisited
child node, mark it as traversed and push it on stack.
Step 5 : If the node does not have any unvisited child nodes, pop the
node from the stack.
2. Eulerian graph : A graph containing an Eulerian circuit is called
Eulerian graph.
For example : Graphs given below are Eulerian graphs.
V1
V2
V5
V3
A
B
V4 D
Fig. 7.
C
Eulerian path : A path of graph G which includes each edge of G
exactly once is called Eulerian path.
Eulerian circuit : A circuit of graph G which include each edge of
G exactly once.
The existence of Eulerian paths or Eulerian circuits in a graph is
related to the degree of vertices.
a. Adjacency matrix :
i. Representation of undirected graph :
The adjacency matrix of a graph G with n vertices and no parallel
edges is a n × n matrix A = [aij] whose elements are given by
aij = 1, if there is an edge between ith and jth vertices
= 0, if there is no edge between them
ii. Representation of directed graph :
The adjacency matrix of a digraph D, with n vertices is the matrix
A = [aij]n×n in which
aij = 1 if arc (vi, vj) is in D
= 0 otherwise
For example :
V1
V4
V2
V3
Fig. 8.
SP–14 C (CS/IT-Sem-3)
Solved Paper (2017-18)
v1
v2
v1 0
A = v2 1
v3 1
v4 1
v3
v4
1 1 1
0 1 0
1 0 1
0 1 0
b. Solve the recurrence relation : ar + 4ar – 2 + 4ar – 2 = r2.
Ans. ar + 4ar – 1 + 4ar – 2 = r2
The characteristic equation is,
x2 + 4x + 4 = 0
(x + 2)2 = 0
x = –2, – 2
The homogeneous solution is, a(h) = (A0 + A1r) (– 2)r
The particular solution be, a(p) = (A0 + A1r) r2
Put ar, ar – 1 and ar – 2 from a(p) in the given equation, we get
r2A0 + A1r3 + 4A0(r – 1)2 + 4A1(r – 1)3 + 4A0(r – 2)2 + 4A1(r – 2)3 = r2
A0(r2 + 4r2 – 8r + 4 + 4r2 – 16r + 16) +A1(r3 + 4r3 – 4 – 12r2 + 12r + 4r3
– 32 – 24r2 + 48r) = r2
A0(9r2 – 24r + 20) + A1(9r3 – 48r2 + 60r – 36) = r2
Comparing the coefficient of same power of r, we get
9A0 – 48A1 = 1
20A0 – 36A1 = 0
...(1)
...(2)
3
5
Solving equation (1) and (2) A0 =
A1
53
159
The complete solution is,
3 5 2
ar = ar(p) + ar(h) = (A0 + A1r) (– 2)r +
r r
53 159
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B.Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2018-19
DISCRETE STRUCTURES AND
THEORY OF LOGIC
Time : 3 Hours
Note : 1.
2.
Max. Marks : 70
Attempt all Sections. If require any missing data; then choose
suitably.
Any special paper specific instructions.
SECTION – A
1. Attempt all questions in brief.
(2 × 7 = 14)
a. Find the power set of each of these sets, where a and b are
distinct elements.
i. {a}
ii. {a, b}
iii. {, {}}
iv. {a, {a}}
b. Define ring and field.
c. Draw the Hasse diagram of D30.
d. What are the contrapositive, converse, and the inverse of
the conditional statement : “The home team wins whenever
it is raining” ?
e. How many bit strings of length eight either start with a ‘1’
bit or end with the two bit ‘00’ ?
f. Define injective, surjective and bijective function.
g. Show that (p q) and p q are logically equivalent.
SECTION-B
2. Attempt any three of the following :
(7 × 3 = 21)
a. A total of 1232 student have taken a course in Spanish, 879
have taken a course in French, and 114 have taken a course
in Russian. Further 103 have taken courses in both Spanish
and French, 23 have taken courses in both Spanish and
Russian, and 14 have taken courses in both French and
Russian. If 2092 students have taken least one of Spanish,
SP–2 C (CS/IT-Sem-3)
Solved Paper (2018-19)
French and Russian, how many students have taken a
course in all three languages ?
b. i. Let H be a subgroup of a finite group G. Prove that order of
H is a divisor of order of G.
ii. Prove that every group of prime order is cyclic.
c. Define a lattice. For any a, b, c, d in a lattice (A, ) if a b and
c d then show that a c b d and a c b d.
d. Show that ((p q) ~ (~p (~q ~r))) (~p ~q) (~ p r) is
a tautology without using truth table.
e. Define a binary tree. A binary tree has 11 nodes. It’s inorder
and preorder traversals node sequences are :
Preorder : A B D H I E J L C F G
Inorder : H D I B J E K A F C G
Draw the tree.
3. Attempt any one part of the following :
(7 × 1 = 7)
a. Prove that if n is a positive integer, then 133 divides 11n + 1 +
122n – 1.
b. Let n be a positive integer and S a set of strings. Suppose
that Rn is the relation on S such that sRnt if and only if
s = t, or both s and t have at least n characters and first n
characters of s and t are the same. That is, a string of fewer
than n characters is related only to itself; a string s with at
least n characters is related to a string t if and only if t has
at least n characters and t beings with the n characters at
the start of s.
4. Attempt any one part of the following :
(7 × 1 = 7)
a. Let G = {1, – 1, i, – i} with the binary operation multiplication
be an algebraic structure, where i =
whether G is an abelian or not.
1 . Determine
b. What is meant by ring ? Give examples of both commutative
and non-commutative rings.
5. Attempt any one part of the following :
(7 × 1 = 7)
a. Show that the inclusion relation is a partial ordering on
the power set of a set S. Draw the Hasse diagram for
inclusion on the set P(S), where S = {a, b, c, d}. Also determine
whether (P(S), ) is a lattice.
Discrete Structures & Theory of Logic
SP–3 C (CS/IT-Sem-3)
b. Find the Sum-Of-Products and Product-Of-sum expansion
of the Boolean function F(x, y, z) = (x + y) z.
6. Attempt any one part of the following :
(7 × 1 = 7)
a. What is a tautology, contradiction and contingency ? Show
that (p q) ( p r) (q r) is a tautology, contradiction
or contingency.
b. Show that the premises “It is not sunny this afternoon and
it is colder than yesterday,” “We will go swimming only if it
is sunny,” “If we do not go swimming, then we will take a
canoe trip.” and “If we take a canoe trip, then we will be
home by sunset” lead to the conclusion “We will be home by
sunset.”
7. Attempt any one part of the following :
(7 × 1 = 7)
a. What are different ways to represent a graph. Define Euler
circuit and Euler graph. Give necessary and sufficient
conditions for Euler circuits and paths.
b. Suppose that a valid codeword is an n-digit number in
decimal notation containing an even number of 0’s. Let an
denote the number of valid codewords of length n satisfying
the recurrence relation an = 8an – 1 + 10n – 1 and the initial
condition a1 = 9. Use generating functions to find an explicit
formula for an.
SP–4 C (CS/IT-Sem-3)
Solved Paper (2018-19)
SOLUTION OF PAPER (2018-19)
Note : 1.
2.
Attempt all Sections. If require any missing data; then choose
suitably.
Any special paper specific instructions.
SECTION – A
1. Attempt all questions in brief.
(2 × 7 = 14)
a. Find the power set of each of these sets, where a and b are
distinct elements.
i. {a}
ii. {a, b}
iii. {, {}}
iv. {a, {a}}
Ans.
i.
ii.
iii.
iv.
Power set of {a} = {{}, {a}}
Power set of {a, b} = {{}, {a}, {b}, {a, b}}
Power set of {, {}} = {}
Power set of {a, {a}} = {{}, {a}, {{a}}, {a, {a}}}
b. Define ring and field.
Ans. Ring : A non-empty set R is a ring if it is equipped with two binary
i.
ii.
iii.
iv.
v.
vi.
i.
ii.
iii.
operations called addition and multiplication and denoted by '+'
and '.' respectively i.e., for all a, b R we have a + b R and a.b R
and it satisfies the following properties :
Addition is associative i.e.,
(a + b) + c = a + (b + c) a, b, c R
Addition is commutative i.e.,
a + b = b + a a, b R
There exists an element 0 R such that
0 + a = a = a + 0, a R
To each element a in R there exists an element – a in R such that
a + (– a) = 0
Multiplication is associative i.e.,
a.(b.c) = (a.b).c, a b, c R
Multiplication is distributive with respect to addition i.e., for all
a, b, c R,
Field : A ring R with at least two elements is called a field if it has
following properties :
R is commutative
R has unity
R is such that each non-zero element possesses multiplicative
inverse.
c. Draw the Hasse diagram of D30.
SP–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Ans.
30
15
6
10
3
2
5
1
Fig. 1.
d. What are the contrapositive, converse, and the inverse of
the conditional statement : “The home team wins whenever
it is raining” ?
Ans. Given : The home team wins whenever it is raining.
q(conclusion) : The home team wins.
p(hypothesis) : It is raining.
Contrapositive : ~ q ~ p is “if the home team does not win
then it is not raining”.
Converse : q p is “if the home team wins then it is raining”.
Inverse : ~ p ~ q is “if it is not raining then the home team does
not win”.
e. How many bit strings of length eight either start with a ‘1’
bit or end with the two bit ‘00’ ?
Ans.
1. Number of bit strings of length eight that start with a 1 bit :
27 = 128.
2. Number of bit strings of length eight that end with bits 00 : 26 = 64.
3. Number of bit strings of length eight 25 = 32 that start with a 1 bit
and end with bits 00 : 25 = 32
Hence, the number is 128 + 64 – 32 = 160.
f. Define injective, surjective and bijective function.
Ans.
1. One-to-one function (Injective function or injection) : Let
f : X Y then f is called one-to-one function if for distinct elements
of X there are distinct image in Y i.e., f is one-to-one iff
f(x1) = f(x2) implies x1 = x2 x1, x2, X
A
B
f
Fig. 2. One-to-one.
SP–6 C (CS/IT-Sem-3)
Solved Paper (2018-19)
2. Onto function (Surjection or surjective function) : Let
f : X Y then f is called onto function iff for every element y Y
there is an element x X with f(x) = y or f is onto if Range (f) = Y.
X
Y
f
Fig. 3. Onto.
3. One-to-one onto function (Bijective function or bijection) :
A function which is both one-to-one and onto is called one-to-one
onto function or bijective function.
X
1
2
3
Y
f
a
b
c
Fig. 4. One-to-one onto.
g. Show that (p q) and p q are logically equivalent.
Ans. To prove : (p + q) = p.q
To prove the theorem we will show that
(p + q) + p.q = 1
Consider (p + q) + p.q = {(p + q) + p}.{(p + q) + q}
by Distributive law
= {(q + p) + p}.{(p + q) + q}
by Commutative law
= {q + (p + p)}.{p + (q + q)}
by Associative law
= (q + 1).(p + 1)
by Complement law
= 1.1
by Dominance law
=1
...(1)
Also consider
(p + q).pq = pq.(p + q)
by Commutative law
= pq.p + pq.q
by Distributive law
= p.(pq) + p.(q q) by Commutative law
= (p. p).q + p(q. q)
by Associative law
= 0. q + p.0
by Complement law
= q.0 + p.0
by Commutative law
= 0+0
by Dominance law
=0
...(2)
From (1) and (2), we get,
pq is complement of (p + q) i.e., (p + q) = p q.
Discrete Structures & Theory of Logic
SP–7 C (CS/IT-Sem-3)
SECTION-B
2. Attempt any three of the following :
(7 × 3 = 21)
a. A total of 1232 student have taken a course in Spanish, 879
have taken a course in French, and 114 have taken a course
in Russian. Further 103 have taken courses in both Spanish
and French, 23 have taken courses in both Spanish and
Russian, and 14 have taken courses in both French and
Russian. If 2092 students have taken least one of Spanish,
French and Russian, how many students have taken a
course in all three languages ?
Ans. Let S be the set of students who have taken a course in Spanish, F
be the set of students who have taken a course in French, and R be
the set of students who have taken a course in Russian. Then, we
have
|S| = 1232, |F| = 879, |R| = 114, |S F| = 103, |S R| = 23,
|S R| = 14, and |SF R| = 23.
Using the equation
|SF R| = |S|+ |F|+ |R| – |S F| – |S R| – |S R| +
|SF R|,
2092 = 1232 + 879 + 114 – 103 – 23 – 14 +|SF R|,
|SF R| = 7.
|S F R| = ?
|S F| = 103
|S| = 1232
|F| = 879
|S R| = 23
|R| = 114
|F R| = 14
|S F R| = 2092
Fig. 5.
b. i. Let H be a subgroup of a finite group G. Prove that order of
H is a divisor of order of G.
Ans.
1. Let H be any sub-group of order m of a finite group G of order n. Let
us consider the left coset decomposition of G relative to H.
2. We will show that each coset aH consists of m different elements.
Let
H = {h1, h2, ....., hm}
3. Then ah1, ah2, ...., ahm, are the members of aH, all distinct.
SP–8 C (CS/IT-Sem-3)
Solved Paper (2018-19)
For, we have
ahi = ahj hi = hj
by cancellation law in G.
4. Since G is a finite group, the number of distinct left cosets will also
be finite, say k. Hence the total number of elements of all cosets is
km which is equal to the total number of elements of G.
Hence
n = mk
This show that m, the order of H, is a divisor of n, the order of the
group G.
We also find that the index k is also a divisor of the order of the
group.
ii. Prove that every group of prime order is cyclic.
Ans.
1.
2.
3.
4.
5.
Let G be a group whose order is a prime p.
Since P > 1, there is an element a G such that a e.
The group <a> generated by ‘a’ is a subgroup of G.
By Lagrange’s theorem, the order of ‘a’ divides |G|.
But the only divisors of |G|= p are 1 and p. Since a e we have
|<a>|> 1, so |<a>| = p.
6. Hence, <a> = G and G is cyclic.
c. Define a lattice. For any a, b, c, d in a lattice (A, ) if a b and
c d then show that a c b d and a c b d.
Ans. Lattice : A lattice is a poset (L, ) in which every subset {a, b}
consisting of 2 elements has least upper bound (lub) and greatest
lower bound (glb). Least upper bound of {a, b} is denoted by a b
and is known as join of a and b. Greatest lower bound of {a, b} is
denoted by a b and is known as meet of a and b.
Lattice is generally denoted by (L, , ).
Numerical :
As a b and c d, a b b d and c d b d.
By transtivity of , a b d and c b d.
So b d is an upper bound of a and c.
So a c bd.
As a c a and a c c, a c a b and a c c d.
Hence a c is a lower bound of b and d. So a c b d.
So a c b d.
d. Show that ((p q) ~ (~p (~q ~r))) (~p ~q) (~ p r) is
a tautology without using truth table.
Ans.
i. We have
((p q) ~ (~ p (~ q r))) (~ p ~ q) (~ p r)
((p q) ~ (~ p ~ (q r))) (~ (p q) ~ ( p r))
SP–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
(Using De Morgan’s Law)
[(p q)] (p (q r)) ~ ((p q) (p r))
[(p q) (p q) (p r)] ~ (( p q) (p r))
(Using Distributive Law)
[((p q) (p q)] (p r) ~ ((p q) (p r))
((p q) (p r)) ~ ((p q) (p r))
x ~ x where x = (p q) (p r)
T
e. Define a binary tree. A binary tree has 11 nodes. It’s inorder
and preorder traversals node sequences are :
Preorder : A B D H I E J L C F G
Inorder : H D I B J E K A F C G
Draw the tree.
Ans. Binary tree : Binary tree is the tree in which the degree of every
node is less than or equal to 2. A tree consisting of no nodes is also
a binary tree.
Numerical :
Step 1 : In preorder sequence, leftmost element is the root of the
tree. By searching A in ‘Inorder Sequence’ we can find out all the
elements on the left and right sides of ‘A’.
A
FCG
HDIBJEK
Step 2 : We recursively follow the above steps and we get
A
B
C
A
B
HDI JEK
D
C
F
G
H
E
I
J
F
G
K
3. Attempt any one part of the following :
(7 × 1 = 7)
a. Prove that if n is a positive integer, then 133 divides 11n + 1 +
122n – 1.
Ans. We prove this by induction on n.
Base case : For n = 1, 11n+1 + 122n–1 = 112 + 121 = 133 which is
divisible by 133.
Inductive step : Assume that the hypothesis holds for n = k, i.e.,
11k+1 + 122k–1 = 133A for some integer A. Then for n = k + 1,
11n+1 + 122n–1 =11k+1+1 + 122(k+1)–1
=
11k+2 + 122k+1
SP–10 C (CS/IT-Sem-3)
Solved Paper (2018-19)
=11 * 11k+1 + 144 * 122k–1
=11 * 11k+1 + 11 * 122k–1 + 133 * 122k–1
=11[11k+1 + 122k–1] + 133 * 122k–1
=11* 133A + 133 * 122k–1
= 133[11A + 122k–1]
Thus if the hypothesis holds for n = k it also holds for n = k + 1.
Therefore, the statement given in the equation is true.
b. Let n be a positive integer and S a set of strings. Suppose
that Rn is the relation on S such that sRnt if and only if
s = t, or both s and t have at least n characters and first n
characters of s and t are the same. That is, a string of fewer
than n characters is related only to itself; a string s with at
least n characters is related to a string t if and only if t has
at least n characters and t beings with the n characters at
the start of s.
Ans. We have to show that the relation Rn is reflexive, symmetric, and
transitive.
1. Reflexive : The relation Rn is reflexive because s = s, so that sRns
whenever s is a string in S.
2. Symmetric : If sRn t, then either s = t or s and t are both at least n
characters long that begin with the same n characters. This means
that tRn s. We conclude that Rn is symmetric.
3. Transitive : Now suppose that sRn t and tRn u. Then either s = t or
s and t are at least n characters long and s and t begin with the same
n characters, and either t = u or t and u are at least n characters
long and t and u begin with the same n characters. From this, we
can deduce that either s = u or both s and u are n characters long
and s and u begin with the same n characters, i.e., sR n u.
Consequently, Rn is transitive.
4. Attempt any one part of the following :
(7 × 1 = 7)
a. Let G = {1, – 1, i, – i} with the binary operation multiplication
be an algebraic structure, where i =
whether G is an abelian or not.
Ans. The composition table of G is
*
1
–1
i
–i
1
1
–1
i
–i
–1
–1
1
–i
i
i
i
–i
–1
1
–i
–i
i
–1
1
1 . Determine
Discrete Structures & Theory of Logic
SP–11 C (CS/IT-Sem-3)
1. Closure property : Since all the entries of the composition table
are the elements of the given set, the set G is closed under
multiplication.
2. Associativity : The elements of G are complex numbers, and we
know that multiplication of complex numbers is associative.
3. Identity : Here, 1 is the identity element.
4. Inverse : From the composition table, we see that the inverse
elements of 1, – 1, i, – i are 1, – 1, – i, i respectively.
5. Commutativity : The corresponding rows and columns of the
table are identical. Therefore the binary operation is commutative.
Hence, (G, *) is an abelian group.
b. What is meant by ring ? Give examples of both commutative
and non-commutative rings.
Ans. Ring : A ring is an algebraic system (R, +, •) where R is a non
empty set and + and • are two binary operations (which can be
different from addition and multiplication) and if the following
conditions are satisfied :
1. (R, +) is an abelian group.
2. (R, •) is semigroup i.e., (a • b) • c = a • (b • c) a, b, c R.
3. The operation • is distributive over +.
i.e., for any a, b, c R
a • (b + c) = (a • b) + (a • c) or (b + c) • a = (b • a) + (c • a)
Example of commutative ring :
Let a, b R (a + b)2 = (a + b)
(a + b) (a + b) = (a + b)
(a + b)a + (a + b)b = (a + b)
(a2 + ba) + (ab + b2) = (a + b)
(a + ba) + (ab + b) = (a + b)
( a2 = a and b2 = b)
(a + b) + (ba + ab) = (a + b) + 0
ba + ab = 0
a + b = 0 a + b = a + a [being every element of its own additive
inverse]
b=a
ab = ba
R is commutative ring.
Example of non-commutative ring : Consider the set R of 2 × 2
matrix with real element. For A, B, C R
A * (B + C) = (A * B) + (A * C)
also, (A + B) * C = (A * C) + (B * C)
* is distributive over +.
(R, +, *) is a ring.
We know that AB BA, Hence (R, +, *) is non-commutative ring.
5. Attempt any one part of the following :
(7 × 1 = 7)
a. Show that the inclusion relation is a partial ordering on
the power set of a set S. Draw the Hasse diagram for
SP–12 C (CS/IT-Sem-3)
Solved Paper (2018-19)
inclusion on the set P(S), where S = {a, b, c, d}. Also determine
whether (P(S), ) is a lattice.
Ans. Show that the inclusion relation () is a partial ordering on the
power set of a set S.
Reflexivity : A A whenever A is a subset of S.
Antisymmetry : If A and B are positive integers with A B and B
A, then A = B.
Transitivity : If A B and B C, then A C.
Hasse diagram :
{a, b, c, d}
{a, b}
{a, c, d}
{a, b, d}
{a, b, c}
{a, c}
{a, c}
{a}
{b, c, d}
{b, c}
{b, d}
{c}
{d}
{b}
{c, d}
Fig. 6.
(P(S), ) is not a lattice because ({a, b},{b, d}) has no lub and glb.
b. Find the Sum-Of-Products and Product-Of-sum expansion
of the Boolean function F(x, y, z) = (x + y) z.
Ans. F(x, y, z) = (x + y)z
x
y
z
x+y
z
(x + y)z
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
0
0
1
0
0
1
1
1
0
1
1
1
0
0
0
1
0
1
1
1
0
0
1
0
0
0
0
0
0
0
1
0
SP–13 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Sum-Of-Product :
F(x, y, z) = xyz + xyz + xyz
Product-Of-Sum :
F(x, y, z) = (x + y + z)(x + y+ z)(x+ y + z)(x+ y+ z)
(x+ y+ z')
6. Attempt any one part of the following :
(7 × 1 = 7)
a. What is a tautology, contradiction and contingency ? Show
that (p q) ( p r) (q r) is a tautology, contradiction
or contingency.
Ans. Tautology, contradiction and contingency :
1. Tautology : Tautology is defined as a compound proposition that is
always true for all possible truth values of its propositional variables
and it contains T in last column of its truth table.
Propositions like,
i. The doctor is either male or female.
ii. Either it is raining or not.
are always true and are tautologies.
2. Contradiction : Contradiction is defined as a compound
proposition that is always false for all possible truth values of its
propositional variables and it contains F in last column of its truth
table.
Propositions like,
i. x is even and x is odd number.
ii. Tom is good boy and Tom is bad boy.
are always false and are contradiction.
3. Contingency : A proposition which is neither tautology nor
contradiction is called contingency.
Here the last column of truth table contains both T and F.
Proof : ((p q) (~ p r)) (q r)
p
q
F
F
F
F
T
T
T
T
(p q) (~ p r) (A B)
=A
=B
=C
(q r) C D
=D
r
~P
F
F
T
F
T
T
F
F
F
T
T
F
T
T
T
T
T
F
T
T
T
T
T
T
T
T
T
T
T
T
T
T
F
F
F
T
F
T
F
F
F
T
F
T
T
T
T
T
T
F
F
T
F
T
T
T
T
T
F
T
T
T
T
T
So, ((p q) (~ p r)) (q r) is contingency.
SP–14 C (CS/IT-Sem-3)
Solved Paper (2018-19)
b. Show that the premises “It is not sunny this afternoon and
it is colder than yesterday,” “We will go swimming only if it
is sunny,” “If we do not go swimming, then we will take a
canoe trip.” and “If we take a canoe trip, then we will be
home by sunset” lead to the conclusion “We will be home by
sunset.”
Ans.
i. The compound proposition will be : (p q r) s
ii. Let p be the proposition “It is sunny this afternoon”, q be the
proposition “It is colder than yesterday”, r be the proposition “We
will go swimming”, s be the proposition ‘‘We will take a canoe trip’’,
and t be the proposition “We will be home by sunset”.
Then the hypothesis becomes p q, r p, r s, and s t. The
conclusion is simply t.
We construct an argument to show that our hypothesis lead to the
conclusion as follows :
S. No.
Step
Reason
1.
pq
Hypothesis
2.
p
Simplification using step 1
3.
rp
Hypothesis
4.
r
Modus tollens using steps 2 and 3
5.
rs
Hypothesis
6.
s
Modus ponens using steps 4 and 5
7.
st
Hypothesis
8.
t
Modus ponens using steps 6 and 7
7. Attempt any one part of the following :
(7 × 1 = 7)
a. What are different ways to represent a graph. Define Euler
circuit and Euler graph. Give necessary and sufficient
conditions for Euler circuits and paths.
Ans. Representation of graph :
Graph can be represented in following two ways :
1. Matrix representation :
Matrices are commonly used to represent graphs for computer
processing. Advantages of representing the graph in matrix lies in
the fact that many results of matrix algebra can be readily applied
to study the structural properties of graph from an algebraic point
of view.
a. Adjacency matrix :
i. Representation of undirected graph
SP–15 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
ii.
b.
i.
ii.
2.
Representation of directed graph
Incidence matrix :
Representation of undirected graph
Representation of directed graph
Linked representation : In this representation, a list of vertices
adjacent to each vertex is maintained. This representation is also
called adjacency structure representation. In case of a directed
graph, a care has to be taken, according to the direction of an edge,
while placing a vertex in the adjacent structure representation of
another vertex.
Euler circuit and Euler graph :
Eulerian circuit : A circuit of graph G which include each edge of
G exactly once.
Eulerain graph : A graph containing an Eulerian circuit is called
Eulerian graph.
For example : Graphs given below are Eulerian graphs.
V1
V2
V5
V3
A
B
V4
D
C
Fig. 7.
Necessary and sufficient condition for Euler circuits and
paths :
1. A graph has an Euler circuit if and only if the degree of every
vertex is even.
2. A graph has an Euler path if and only if there are at most two
vertices with odd degree.
b. Suppose that a valid codeword is an n-digit number in
decimal notation containing an even number of 0’s. Let an
denote the number of valid codewords of length n satisfying
the recurrence relation an = 8an – 1 + 10n – 1 and the initial
condition a1 = 9. Use generating functions to find an explicit
formula for an.
Ans. Let G(x) =
an xn be the generating function of the sequence a0,
n 0
a1, a2 .... we sum both sides of the last equations starting with n = 1.
To find that
G(x) – 1 =
n
a x (8a
n –1 x
n
n 1
= 8
10 n–1 x n )
a
n –1 x
n 1
n
n 1
n
10
n 1
n –1 n
x
SP–16 C (CS/IT-Sem-3)
Solved Paper (2018-19)
= 8x
a
n –1 x
n –1
x
n 1
n –1 n –1
x
n 1
= 8x
10
a x
n
n
x
n 0
10
n
xn
n 0
= 8xG(x) + x/(1 – 10x)
Therefore, we have
G(x) – 1 = 8xG(x) + x/(1 – 10x)
Expanding the right hand side of the equation into partial fractions
gives
1 1
1
2 1 8 x 1 10 x
This is equivalent to
G(x) =
G(x) =
1
n n
8 x
2 n 0
=
n 0
an =
10
n 0
1 n
(8 10n ) x n
2
1 n
(8 10n )
2
x
n n
Discrete Structures & Theory of Logic
SP–1 C (CS/IT-Sem-3)
B.Tech.
(SEM. III) ODD SEMESTER
THEORY EXAMINATION, 2019-20
DISCRETE STRUCTURES & THEORY
OF LOGIC
Time : 3 Hours
Note : 1.
Max. Marks : 100
Attempt all Section.
Section-A
1. Answer all questions in brief.
a. Define various types of functions.
(2 × 10 = 20)
b. How many symmetric and reflexive relations are possible
from a set A containing ‘n’ elements ?
c. Let Z be the group of integers with binary operation *
defined by a * b = a + b – 2, for all a, b Z. Find the identity
element of the group (Z, *).
d. Show that every cyclic group is abelian.
e. Prove that a lattice with 5 elements is not a boolean
algebra.
f. Write the contrapositive of the implication: “if it is Sunday
then it is a holiday”.
g. Show that the propositions p q and p q are logically
equivalent.
h. Show that there does not exist a graph with 5 vertices
with degress 1, 3, 4, 2, 3 respectively.
i. Obtain the generating function for the sequence 4, 4, 4, 4, 4,
4, 4.
j. Define Pigeon hole principle.
Ans. Refer Q. 5.13, Page SQ–21C, Unit-5, Two Marks Questions.
Section-B
2. Answer any three of the following :
(3 × 10 = 30)
SP–2 C (CS/IT-Sem-3)
a. Prove that
Solved Paper (2019-20)
1
1
+
1
2
+
1
3
+ ............ +
1
n
> n for n 2 using
principle of mathematical induction.
b. What do you mean by cosets of a subgroup ? Consider the
group Z of integers under addition and the subgroup
H = {....., – 12, – 6, 0, 6, 12, .....} considering of multiple of 6
i. Find the cosets of H in Z.
ii. What is the index of H in Z.
c. Show that the following are equivalent in a Boolean
algebra.
a b a * b = 0 b a a b = 1
d. Show that ((P Q) (QR)) (P Q) (P R) is
tautology by using equivalences.
e. Define planar graph. Prove that for any connected planar
graph, v – e + r = 2 where v, e, r is the number of vertices,
edges, and regions of the graph respectively.
Section-C
3. Answer any one part of the following :
(1 × 10 = 10)
a. Find the number between 1 to 500 that are not divisible by
any of the integers 2 or 3 or 5 or 7.
b. Is the “divides” relations on the set of positive integers
transitive ? What is the reflexive and symmetric closure of
the relation ?
R = {(a, b)|a > b} on the set of positive integers.
4. Answer any one part of the following :
(1 × 10 = 10)
a. What is ring ? Define elementary properties of ring with
example.
b. Prove or dis prove that intersection of two normal
subgroups of a group G is again a normal subgroup of G.
5. Answer any one part of the following :
(10 × 1 = 10)
a. Let (L, , , ) be a distribute lattice and a, b L. If a b = a
c and a b = a c then show that b = c.
b. Obtain the principle disjunction and conjunctive normal
forms of the formula (p r) (q p).
Discrete Structures & Theory of Logic
SP–3 C (CS/IT-Sem-3)
6. Answer any one part of the following :
(1 × 10 = 10)
a. Explain various rules of inference for propositional logic.
b. Prove the validity of the following argument “if the races
are fixed so the casinos are cooked, then the tourist trade
will decline. If the tourist trade decreases, then the police
will be happy. The police force is never happy. Therefore,
the races are not fixed.”
7. Answer any one part of the following :
(7 × 1 = 7)
a. Solve the following recurrence equation using generating
function
G(K) – 7 G(K – 1) + 10 G (K – 2) = 8K + 6
b. A collection of 10 electric bulbs contain 3 defective ones.
i. In how many ways can a sample of four bulbs be selected ?
ii. In how many ways can a sample of 4 bulbs be selected
which contain 2 good bulbs and 2 defective ones ?
iii. In how many ways can a sample of 4 bulbs be selected so
that either the sample contain 3 good ones and 1 defectives
ones or 1 good and 3 defectives ones ?
SP–4 C (CS/IT-Sem-3)
Solved Paper (2019-20)
SOLUTION OF PAPER (2019-20)
Note : 1.
Attempt all Section.
Section-A
1. Answer all questions in brief.
(2 × 10 = 20)
a. Define various types of functions.
Ans. Various types of functions :
1. Algebraic functions : Algebraic functions are those functions
which consist of a finite number of terms involving powers and
roots of the independent variable x.
2. Transcendental functions : A function which is not algebraic is
called transcendental function.
b. How many symmetric and reflexive relations are possible
from a set A containing ‘n’ elements ?
Ans. There are 2n(n + 1)/2 symmetric binary relations and 2n(n – 1) reflexive
binary relations are possible on a set S with cardinality.
c. Let Z be the group of integers with binary operation *
defined by a * b = a + b – 2, for all a, b Z. Find the identity
element of the group (Z, *).
Ans. Since Z is closed for addition, as we have
a + b Z for all a, b Z
a+b– 2Z
a* b Z
So * is a binary operation on Z.
Again,
a*b= a+b–2=b+a–2
(by commutative law of addition on Z)
= b * a for all a, b Z
Hence * is commutative.
Again, (a * b) * c = (a + b – 2) * c
= (a + b – 2) + c – 2 = (a + b + c) – 4
and
a * (b * c) = a * (b + c – 2)
= a + (b + c – 2) – 2 = (a + b + c) – 4
Thus, (a * b) * c = a * (b * c) for all a, b, c Z
Hence, * is associative.
Now, if e is the identity element in Z for *, then for all a Z
a* e= aa+e–2=ae=2Z
So, 2 is the identity element for * in Z.
Let the integer a have its inverse b. Then,
a* b=–1 a+b–2=1b=3–a
So, the inverse of a is (3 – a)
d. Show that every cyclic group is abelian.
Ans. Let G be a cyclic group and let a be a generator of G so that
G = <a> = {an : n Z}
SP–5 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
If g1 and g2 are any two elements of G, there exist integers r and s
such that g1 = ar and g2 = as. Then
g1 g2 = ar as = ar+s = as+r = as . ar = g2g1
So, G is abelian.
e. Prove that a lattice with 5 elements is not a boolean
algebra.
Ans. To be a boolean algebra the number of elements in the lattice
should be of the form 2m. Since the number of elements is 5 which
is not of the form 2m. So, it is not a boolean algebra.
f. Write the contrapositive of the implication: “if it is Sunday
then it is a holiday”.
Ans. Consider the statements :
p : It is sunday
q : It is a holiday
Contrapositive : ~ q ~ p
g. Show that the propositions p q and p q are logically
equivalent.
Ans. Truth table :
pq:
p
q
pq
T
T
F
F
p q :
T
F
T
F
T
F
T
T
p
q
p
pq
T
T
F
F
T
F
T
F
F
F
T
T
F
F
T
F
No, both are not equivalent.
h. Show that there does not exist a graph with 5 vertices
with degress 1, 3, 4, 2, 3 respectively.
Ans. According to hand shaking theorem, sum of degree of all the
vertices is even. But in the given degree sequence
1 + 3 + 4 + 2 + 3 = 13
Sum of degrees of all vertices is 13 which is an odd number.
Hence, no such graph exists with the given degree sequence.
i. Obtain the generating function for the sequence 4, 4, 4, 4, 4,
4, 4.
SP–6 C (CS/IT-Sem-3)
Solved Paper (2019-20)
Ans. Here
a0 = 4, a1 = 4, a2 = 4,
G(x) = 4 + 4x + 4x2 + 4x3 + ......
= 4(1 + x + x2 + x3 + ......)
which is an infinite G.P.
G(x) =
4
1 x
j. Define Pigeon hole principle.
Ans. Pigeonhole principle : If n pigeons are assigned to m pigeonholes
then at least one pigeon hole contains two or more pigeons (m < n).
Section-B
2. Answer any three of the following :
a. Prove that
1
1
+
1
+
2
1
3
+ ............ +
(3 × 10 = 30)
1
n
> n for n 2 using
principle of mathematical induction.
1
1
1
.....
n ,n2
1
2
n
Step I : Inductive base :
Ans. Let
S(n) :
S(2) :
1
1
1
2
1.707 1.414 2
S(1) is true.
Step II : Inductive hypothesis : Let us assume S(k) is true.
1
1
1
....
k
1
2
k
Step III : Inductive step : We have to show S(k + 1) is true
i.e.,
i.e.,
1
1
1
....
1
2
k
1
k 1
k1
Consider
1
1
1
....
1
2
k
1
>
k1
k
1
k1
[Using inductive hypothesis]
k k1 1
k k 1
k1
k1
k1
k 1
k1
S(k + 1) is true
Hence by principle of mathematical induction, S(n) is true for all
n N.
=
b. What do you mean by cosets of a subgroup ? Consider the
group Z of integers under addition and the subgroup
H = {....., – 12, – 6, 0, 6, 12, .....} considering of multiple of 6
Discrete Structures & Theory of Logic
SP–7 C (CS/IT-Sem-3)
i. Find the cosets of H in Z.
ii. What is the index of H in Z.
Ans. Coset : Let H be a subgroup of group G and let a G then the set
Ha = {ha : h H} is called right coset generated by H and a.
Also the set aH = {ah : h H} is called left coset generated by a and
H.
Numerical :
i. We have Z = {– 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4, 5, 6, ....}
and
H = {...., – 12, – 6, 0, 6, 12, ....}
Let 0 Z and the right cosets are given as
H = H + 0 = {...., – 12, – 6, 0, 6, 12, ....}
H + 1 = {...., – 11, – 5, 1, 7, 13, ....}
H + 2 = {...., – 10, – 4, 2, 8, 14, ....}
H + 3 = {...., – 9, – 3, 3, 9, 15, ....}
H + 4 = {....., – 8, – 2, 4, 10, 16, ....}
H + 5 = {...., – 7, – 1, 5, 11, 17, ....}
H = H + 6 = {...., – 6, 0, 6, 12, 18, ....}
Now, its repetition starts. Now, we see that the right cosets,
H, H + 1, H + 2, H + 3, H + 4, H + 5 are all distinct and more over they
are disjoint. Similarly the left cosets will be same as right cosets.
ii. Index of H in Z is the number of distinct right/left cosets.
Therefore, index is 6.
c. Show that the following are equivalent in a Boolean
algebra.
a b a * b = 0 b a a b = 1
Ans. First prove the LHS, for this we use a * b = 0 or a b = 0 (say)
Now, a = a I
[By identity of bounded lattice]
= a (b b) = (a b) (a b)
[By law of distribution]
= (a b) 0
[a b = 0]
= abb
[Partial order relation]
Hence it is clear that if a * b = 0 then relation a b is a partial order
relation.
Since, it is evident in Boolean algebra for any a, b,
a b iff b a
[By law of duality]
Note : In the given question, the equivalency cannot be solved if
we use a b = 1. So, assuming a + b = 1.
Since, a b so we can say that a b = a a b = b
If a + b = 1 then by DeMorgan’s law and involution,
1 = 0 = (a * b) = a + b = a + b
0 = 1 = (a+ b) = a * b = a * b
So, a * b = 0 is equivalent to a + b = 1
d. Show that ((P Q) (QR)) (P Q) (P R) is
tautology by using equivalences.
Ans. = ((P Q) ( Q R)) ( P Q) ( P R)
= ((P Q) (Q R)) (( P Q) ( P R))
= ((P Q R) (Q Q R)) (( P P R) ( Q P R))
= (P Q R) (Q R) ( P R) ( Q P R)
SP–8 C (CS/IT-Sem-3)
Solved Paper (2019-20)
= (P T) (Q R) (T Q)( P R)
= (Q R) ( P R)
So, the given expression is not a tautology.
[ T P = T]
e. Define planar graph. Prove that for any connected planar
graph, v – e + r = 2 where v, e, r is the number of vertices,
edges, and regions of the graph respectively.
Ans. Planar graph :
A graph G is said to be planar if there exists some geometric
representation of G which can be drawn on a plane such that no
two of its edges intersect except only at the common vertex.
i. A graph is said a planar graph, if it cannot be drawn on a plane
without a crossover between its edges crossing.
ii. The graphs shown in Fig. 2(a) and (b) are planar graphs.
(a)
Fig. 1. Some planar graph.
(b )
Proof : We will use induction to prove this theorem.
Step I : Inductive base :
Assume that e = 1 Then we have two cases given in figure below :
(a)
(b)
Fig. 2.
In Fig. 1 (a) we have v = 2 and r = 1 v + r – e = 2 + 1 – 1 = 2
In Fig. 1 (b) we have v = 1 and r = 2 v + r – e = 1 + 2 – 1 = 2
Hence vertified
Step II : Inductive hypothesis :
Let us assume that given theorem is true for e = k
i.e., for k edges
Step III : Inductive step :
We have to show that theorem is true for k + 1 edges.
Let graph G has k + 1 edges.
Case I : We suppose that g contain no circuits. Now take a vertex
v and find a path starting at v. Since g has no circuit so whenever
we find an edge we have a new vertex atleast we will reach a
vertex with degree one as shown in Fig. 3.
x
Fig. 3.
v
SP–9 C (CS/IT-Sem-3)
Discrete Structures & Theory of Logic
Now remove vertex x and edge incident on v.
Then we will left with graph G* given as
Fig. 4.
Therefore Euler’s formula holds for graphs in Fig. 4, since it has k
edges [By inductive hypothesis]
Since G has one more edge than G* and one more vertices than G*.
So, let v = v1 + 1 and e = e1 + 1 where G* = (v1, e1)
v + r – e = v1 + 1 + r – e1 – 1
= v1 + r – e1 = 2
[By inductive hypothesis]
Hence Euler’s formula holds true.
Case II : We assume that G has a circuit and e is edge in circuit.
Let G be given in Fig. 5.
v1
v4
e
v3
v2
Fig. 5.
Now e is the part of boundary for 2 region so after removing edge
we are left with graph G* as shown in Fig. 6.
v1
v4
v2
v3
Fig. 6.
Now number of edges in G* are k so by inductive hypothesis,
Euler formula holds for G*.
Now since G has one more edges and region than G* with same vertices.
So
v + r – e = v + r1 + 1 – e1 – 1 = v + r1 – e1 = 2
Hence Euler’s formula also holds for G.
Hence by Principle of mathematical induction Euler’s Theorem
holds true.
Section-C
3. Answer any one part of the following :
(1 × 10 = 10)
a. Find the number between 1 to 500 that are not divisible by
any of the integers 2 or 3 or 5 or 7.
Ans. Let A, B, C and D be the sets of integers between 1 and
500(inclusive) which are divisible by 2, 3, 5, and 7, respectively.
We want |(A B C D)C|. Now
SP–10 C (CS/IT-Sem-3)
Solved Paper (2019-20)
|(A B C D)| = |A| + |B| + |C| + |D| – |A B| – |A
C| – |A D| – |B C| – |B D| – |C D| + |A B C| +
|A B D| + |A C D| + |B C D| – |A B C D|
We have
500
500
500
500
|A| =
= 250 |B| =
= 166 |C| =
= 100 |D| =
3
5
7
2
= 71
500
500
500
|A B| =
= 83 |A C| =
= 50 |A D| =
= 35
6
10
14
500
500
500
|B C| =
= 33 |B D| =
= 23 |C D| =
= 14
15
21
35
500
500
|A B C| =
= 16 |A B D| =
= 11
30
42
500
500
|A C D| =
= 7 |B C D| =
=4
70
105
500
|A B C D| =
=2
210
So |A B C D| = 385.
Hence, the number between 1 to 500 that are not divisible by 2 or
3 or 5 or 7 is 500 – 385 = 115.
b. Is the “divides” relations on the set of positive integers
transitive ? What is the reflexive and symmetric closure of
the relation ?
R = {(a, b)|a > b} on the set of positive integers.
Ans. Yes, divides relation on the set of positive integers is transitive.
Numerical :
Reflexive :
a = a aa
(a, a) R a is real number.
R is reflexive
Symmetric : Let (a, b) R
a b b a (b, a) R
R is not symmetric
R is not an equivalence relation.
4. Answer any one part of the following :
(1 × 10 = 10)
a. What is ring ? Define elementary properties of ring with
example.
Ans. Ring : A ring is an algebraic system (R, +, •) where R is a nonempty set and + and • are two binary operations (which can be
different from addition and multiplication) and if the following
conditions are satisfied :
1. (R, +) is an abelian group.
2. (R, •) is semigroup i.e., (a • b) • c = a • (b • c) a, b, c R.
Discrete Structures & Theory of Logic
SP–11 C (CS/IT-Sem-3)
3. The operation • is distributive over +.
i.e., for any a, b, c R
a • (b + c) = (a • b) + (a • c) or (b + c) • a = (b • a) + (c • a)
Elementary properties of a ring :
Let a, b and c belong to a ring R. Then
1. a.0 = 0. a = 0
2. a.(– b) = (– a).b = – (a.b)
3. (– a).(– b) = a.b
4. a(b – c) = a.b – a.c and (b – c).a = b.a – c.a
For example : If a, b R then (a + b)2 = a2 + ab + ba + b2
We have (a + b)2 = (a + b) (a + b)
= a (a + b) + b (a + b) [by right distributive law]
= (aa + ab) + (ba + bb)[by right distributive law]
= a2 + ab + ba + b2
b. Prove or dis prove that intersection of two normal
subgroups of a group G is again a normal subgroup of G.
Ans. Let H1 and H2 be any two subgroups of G. Since at least the identity
element e is common to both H1 and H2.
H1 H2
In order to prove that H1 H2 is a subgroup, it is sufficient to prove
that
a H1 H2, b H1 H2 ab–1 H1 H2
Now a H1 H2 a H1 and a H2
b H1 H2 b H1 and b H2
But H1, H2 are subgroups. Therefore,
a H1, b H1 ab–1 H1
a H2 , bH2 ab–1 H2
Finally, ab–1 H1, ab–1 H2 ab–1 H1 H2
Thus, we have shown that
a H1 H2, b H1 H2 ab–1 H1 H2.
Hence, H1 H2 is a subgroup of G.
5. Answer any one part of the following :
(10 × 1 = 10)
a. Let (L, , , ) be a distribute lattice and a, b L. If a b = a
c and a b = a c then show that b = c.
b = b (a b)
(Absorption)
Ans.
= b (a c)
(hypothesis)
= (b a) (b c)
(distributive law)
= (a c) (b c)
(hypothesis)
= (a b) c
(distributive law)
= (a c) c
(hypothesis)
=c
(Absorption)
b. Obtain the principle disjunction and conjunctive normal
forms of the formula (p r) (q p).
Ans. Principle disjunction normal form :
(p r) (p q)
SP–12 C (CS/IT-Sem-3)
Solved Paper (2019-20)
(~ p r) ((p q) (p q))
[By using distributive property]
B C
A
(B A) (C A)
A = (q p) (p q) = (~ q p) (~ p q)
= (~ q p) ~ p (~ q p) q
= (~ q ~ p) (p p) (~ q ~ q) (p q)
[By using distributive property]
= (~ q ~ p) F F (p q) = (~ q ~ p) (p q)
Now, (B A) (C A)
= ~ p ((~ q ~ p) (p q)) (r (~ q ~ p) (p q))
= (~ p ~ q ~ p) (~ p p q)) (r ~ q ~ p) (r p q))
= (~ q ~ p) (r ~ q ~ p) (r p q)
= (T r)(~ q ~ p) (r p q) = (~ q ~ p) (r p q)
Principle conjunctive normal form :
(p r) (p q)
(p r) (q p) (p q)
(~ p r) (~ q p) (~ p q)
6. Answer any one part of the following :
(1 × 10 = 10)
a. Explain various rules of inference for propositional logic.
Ans. Rules of inference are the laws of logic which are used to reach
the given conclusion without using truth table. Any conclusion
which can be derived using these laws is called valid conclusion
and hence the given argument is valid argument.
1. Modus ponens (Law of detachment) : By this rule if an
implication p q is true and the premise p is true then we can
always conclude that q is also true.
The argument is of the form :
p q
p
q
2. Modus tollens (Law of contraposition) : By this rule if an
implication p q is true and conclusion q is false then the premise
p must be false.
The argument is of the form :
p q
~q
~p
3. Hypothetical syllogism : By this rule whenever the two
implications p q and q r are true then the implication p r is
also true.
The argument is of the form :
Discrete Structures & Theory of Logic
SP–13 C (CS/IT-Sem-3)
p q
q r
p r
4. Disjunctive syllogism : By this rule if the premises p q and ~ q
are true then p is true.
The argument is of the form :
p q
~q
p
5. Addition : By this rule if p is true then p q is true regardless the
truth value of q.
The argument is of the form :
p
p q
6. Simplification : By this rule if p q is true then p is true.
The argument is of form :
p q
p q
p or q
7. Conjunction : By this rule if p and q are true then p q is true.
The argument is of the form :
p
q
p q
8. Constructive dilemma : By this rule if (p q) (r s) and p r
are true then q s is true.
The argument is of form :
( p q) ( r s)
p r
q s
9. Destructive dilemma : By this rule if (p q) (r s) and ~ q
s are true.
The argument is of the form :
( p q) (r s)
~qs
pr
10. Absorption : By this rule if p q is true then p (p q) is true.
The argument is of the form :
p q
p ( p q)
SP–14 C (CS/IT-Sem-3)
Solved Paper (2019-20)
b. Prove the validity of the following argument “if the races
are fixed so the casinos are cooked, then the tourist trade
will decline. If the tourist trade decreases, then the police
will be happy. The police force is never happy. Therefore,
the races are not fixed.”
Ans. Let
p : Race are fixed.
q : Casinos are cooked.
r : Tourist trade will decline.
s : Police will be happy.
The above argument can be written in symbolic form as
(p q) r
r s~s
~ p
So,
1. (p q) r (Given Premise) 2. r s (Given Premise)
3. ~ s (Given Permise)
4. ~ r Modus tollens using 2 and 3
5. ~ (p q) Modus tollen using 1 and 4
6. ~ p ~ q
7. Answer any one part of the following :
(7 × 1 = 7)
a. Solve the following recurrence equation using generating
function
G(K) – 7 G(K – 1) + 10 G (K – 2) = 8K + 6
Ans. G(K) – 7 G(K – 1) + 10 G (K – 2) = 8K + 6
Now, the characteristics equation is :
x2 – 7x + 10 = 0
x2 – 5x – 2x + 10 = 0
x(x – 5) – 2(x – 5) = 0
x = 2, 5
The homogeneous solution is Gh(K) = C12K + C25K
Let the particular solution be
Gp(K) = A0 + A1K
A0 + A1 K – 7{A0 + A1(K – 1)} + 10 {A0 + A1(K – 2)} = 8K + 6
A0 + A1 K – 7A0 – 7A1K + 7A1 + 10A0 + 10A1K – 20A1 = 8K + 6
4 A0 + 4A1K – 13A1 = 8K + 6
Comparing the coefficient we get,
4A1 = 8
A1 = 2
4 A0 – 13A1 = 6
A0 = (6 + 13A1)/4 = (6 + 26)/4 = 8
Solution is given by
= Gh(K) + Gp(K) = C12K + C22K + 2K + 8
Discrete Structures & Theory of Logic
SP–15 C (CS/IT-Sem-3)
b. A collection of 10 electric bulbs contain 3 defective ones.
i. In how many ways can a sample of four bulbs be selected ?
ii. In how many ways can a sample of 4 bulbs be selected
which contain 2 good bulbs and 2 defective ones ?
iii. In how many ways can a sample of 4 bulbs be selected so
that either the sample contain 3 good ones and 1 defectives
ones or 1 good and 3 defectives ones ?
Ans.
i. Four bulbs can be selected out of 10 bulbs in
10! 10 9 8 7
= 210 ways
4! 6! 4 3 2 1
ii. Two bulbs can be selected out of 7 good bulbs in 7C2 ways and 2
defective bulbs can be selected out of 3 defective bulbs in 3C2 ways.
Thus, the number of ways in which a sample of 4 bulbs containing
2 good bulbs and 2 defective bulbs can be selected as
10C =
4
7!
3!
76
3 = 63
2! 5! 2! 1!
2
iii. Three godd bulbs can be selected from 7 good bulbs in 7C3 ways
and 1 defective bulb can be selected out of 3 defective ones in 3C1
ways.
Similarly, one good bulb can be selected from 7 good bulb in 7C1
ways and 3 defective ones in 3C3 ways.
So, the number of ways of selecting a sample of 4 bulbs containing
3 good ones and 1 defective or 1 good and 3 defective ones are
7C × 3C =
2
2
7C
3
× 3C1 + 7C1 × 3C3 =
=
7!
3!
7!
3!
3! 4! 1! 2! 1! 6! 3! 0!
765
3 7 = 35 × 3 + 7 = 112
32
